Glass 




Book_. 



X 



WENTWORTEPS 
SERIES OF MATHEMATICS. 



Primary School Arithmetic. 

Grammar School Arithmetic. 

Practical Arithmetic. 

Practical Arithmetic (Abridged Edition), 

Exercises in Arithmetic. 

Shorter Course in Algebra, 

Elements of Algebra. 

Complete Algebra. 

University Algebra. 

Exercises in Algebra. 

Plane Geometry. 

Plane and Solid Geometry. 

Exercises in Geometry. 

PI. and Sol. Geom. and PI. Trigonometry. 

Plane Trigonometry and Tables. 

PI. and Sph. Trig., Surveying, and Tables. 

Trigonometry, Surveying, and Navigation. 

Log. and Trig. Tables (Seven). 

Log. and Trig. Tables (Complete Edition). 



Special Circular and Terms on application. 



ELEMENTS 



9-fL 

ate* 



Analytic Geometry. 



BY 



G. A. WENTWORTH, A.M., 

PROFESSOR OF MATHBMATKS IN PH,LI.I1>S EXETER ACADEMY. 



<*Ko 





BOSTON: 

PUBLISHED BY GINN & COMPANY. 

1886. 



Entered, according to the Act of Congress, in the year 1S86, by 
G. A. WENTWORTH, 

in the Office of the Librarian of Congress, at Washington. 



J. S. Cushing & Co., Printers, Boston. 



PREFACE. 



rriHIS book is intended for beginners. As beginners generally find 
great difficulty in comprehending the connection between a locus 
and its equation, the opening chapter is devoted mainly to an attempt, 
by means of easy illustrations and examples, to make this connection 
clear. 

Each chapter abounds in exercises ; for it is only by solving 
problems which require some degree of original thought that any 
real mastery of the study can be gained. 

The more difficult propositions have been put at the ends of the 
chapters, under the heading of " Supplementary Propositions." This 
arrangement makes it possible for every teacher to mark out his own 
course. The simplest course will be Chapters I.— III. and Chapters 
V.-VIL, with Review Exercises and Supplementary Propositions left 
out. Between this course and the entire work the teacher can exer- 
cise his choice, and take just so much as time and circumstances will 
allow. 

The author has gathered his materials from many sources, but he 
is particularly indebted to the English treatise of Chakles Smith. 
Special acknowledgment is due to G. A. Hill, A.M., of Cambridge, 
Mass., for assistance in the preparation of the book. 

Corrections and suggestions will be thankfully received. 

G. A. WENTWORTH. 
Phillips Exeter Academy, 
July, 1886. 



CONTENTS. 



CHAPTER I. Loci and their Equations. 

SECTIONS. PAGE. 

1-4. Preliminary Notions 1 

5. Circular Measure of an Angle 5 

6-7. Distance between Two Points 6 

8-9. Division of a Line 8 

10-16. Constants and Variables 10 

17-24. Locus of an Equation 14 

25-29. Intersections of Loci 22 

30-34. Construction of Loci 25 

35. Equation of a Curve 31 

Review Exercises 33 



CHAPTER II. The Straight Line. 

36-41. Equations of the Straight Line 36 

42-47. General Equation of the First Degree 43 

48-51. Parallels and Perpendiculars 47 

52-53. Angles 50 

54-56. Distances 54 

57-58. Areas 59 

Review Exercises 62 

59-67. Supplementary Propositions 67 



CHAPTER III. The Circle. 

68-70. Equations of the Circle 77 

71-75. Tangents and Normals 83 

Review Exercises 90 

76-87. Supplementary Propositions 95 



VI CONTENTS. 



CHAPTER IV. Different Systems of Co-ordinates. 

SECTIONS. PAGE. 

88-90. Oblique Co-ordinates 105 

91-92. Polar Co-ordinates 107 

93-100. Transformation of Co-ordinates Ill 

Review Exercises 117 



CHAPTER V. The Parabola. 

101-108. The Equation of the Parabola 119 

109-112. Tangents and Normals 126 

Review Exercises 131 

113-127. Supplementary Propositions 134 



CHAPTER VI. • The Ellipse. 

128-140. Simple Properties of the Ellipse 145 

141-147. Tangents and Normals 155 

Review Exercises 161 

148-164. Supplementary Propositions 163 

CHAPTER VII. The Hyperbola. 

165-173. Simple Properties of the Hyperbola .180 

174-179. Tangents and Normals 187 

Review Exercises 189 

180-197. Supplementary Propositions 190 

CHAPTER VIII. Loci of the Second Order. 

198-200. General Equation of the Second Degree ...... 201 

201-202. Loci of the Second Order that have a Finite Centre . . 207 

203. Loci of the Second Order that have no Finite Centre . . 210 

204. Special Cases of Loci 215 

205. Summary 217 

206-208. Examples and Definitions 217 

Exercises 221 



ANALYTIC GEOMETRY. 



CHAPTER I. 
LOCI AND THEIR EQUATIONS. 

Preliminary Notions. 

1. Let XX f and YY 1 (Fig. 1) be two fixed lines perpen- 
dicular to each other, and intersecting in a point 0. These 
lines divide the plane in which they lie into four similar parts. 





D 


F 




B 


E 


n 




P r F 


x! 


N 





M 


X 


G ' 


B _... 

C 


T 





A 2 



Fig. I. 

Let these parts be called Quadrants (as in Trigonometry), and 
distinguished by naming the area between OX and OY the 
first quadrant ; that between Y and OX' the second quad- 
rant ; that between OX 1 and Y' the third quadrant ; and 
that between OY' and OX the fourth quadrant. 

Suppose the position of a point is described by saying that its 



2 ANALYTIC GEOMETRY. 

distance from YY', expressed in terms of some chosen unit 
of length, is 3, and its distance from XX* is 4. It is clear 
that in each quadrant there is one point, and only one, which 
will satisfy these conditions. The position of the point in 
each quadrant may be found by drawing parallels to YY 1 
at the distance 3 from Y Y\ and parallels to XX* at the dis- 
tance 4 from XX' ; then the intersections P 1? P 2 , P3, and P 4 
satisfy the given conditions. 





p 


.r 




B 


E~" 


"\r 






Im* 


X! 


\jsr 





M 


X 




i& 


T 




Pi 










G 


10 


Y>~ 




A H 



Fig. I. 



2. In order to determine which one of the four points, 
P u P 2 , P 3 , Pi, is meant, we adopt the rule that opposite 
directions shall be indicated by unlike signs. As in Trigo- 
nometry, distances measured from YY 1 to the right are con- 
sidered positive; to the left, negative. Distances measured from 
XX 1 upward are positive; downward, negative. Then the 
position of P 1 will be denoted by +3, +4 ; of P 2 , by — 3, +4 ; 
of P 3 , by -3, -4; of P 4 , by +3, -4. 



3. This method of determining the position of a point in a 
plane is the method commonly employed in Analytic Geom- 
etry. It enables us to represent position by numbers ; and 
by reasoning with these numbers, to investigate the properties 
of geometrical figures. The science of Analytic Geometry 
consists of investigations of this kind. 



LOCI AND THEIR EQUATIONS. 3 

Note. The first man who employed this method successfully in 
investigating the properties of certain figures was the French philoso- 
pher Descartes, whose work on Geometry appeared in the year 1637. 

4. The fixed lines XX' and YY' are called the Axes of 
Oo-ordinates ; XX f is called the Axis of Abscissas, or Axis of x ; 
YY\ the Axis of Ordinates, or Axis of y. The intersection is 
called the Origin. 

The two distances (with signs prefixed) which determine 
the position of a point are called the Oo-ordinates of the point ; 
the distance of the point from YY' is called its Abscissa; and 
the distance from XX f , its Ordinate. 

The letters x and y are in common use as general symbols 
or abbreviations for the words " abscissa" and "ordinate" 
respectively. For the sake of brevity, a point is often rep- 
resented algebraically by simply writing the values of its 
co-ordinates within brackets, the value of the abscissa being 
always written first. 

Thus P l (Fig. 1) is the point (3, 4), P 2 the point (-3, 4), 
P 3 the point (—3, —4), and P 4 the point (3, —4). In gen- 
eral the point whose co-ordinates are x and y is the point 
(x,y). 

Ex. 1. 

1. What are the co-ordinates of the origin ? 

2. In what quadrants are the following points (a and b 
being given lengths) : 

(—a,—b), (—a, J), (a, b), (a, -b). 

3. To what quadrants is a point limited if its abscissa is 
positive ? negative ? ordinate positive ? ordinate negative ? 

4. In what line does a point lie if its abscissa = ? if its 
ordinate = ? 

5. A point (#, y) moves parallel to the axis of x ; which one 
of its co-ordinates remains constant in value ? 



4 ANALYTIC GEOMETRY. 

6. Construct or plot the points : (2, 3), (3, -3), (-1, -3), 
(-4,4), (3,0), (-3,0), (0,4), (0,-1), (0,0). 

Note. To plot a point is to mark its proper position on paper, when 
its co-ordinates are given. The first thing to do is to draw the two 
axes. The rest of the work is obvious after a study of Nos. 1-3. 

7. Construct the triangle whose vertices are the points 
(2,4), (-2,7), (-6,-8). 

8. Construct the quadrilateral whose vertices are the points 
(7,2), (0,-9), (-3,-1), (-6,4). 

9. Construct the quadrilateral whose vertices are (—3,6), 
(-3, 0), (3, 0), (3, 6). What kind of a quadrilateral is it ? 

10. Mark the four points (2, 1), (4, 3), (2, 5), and (0, 3), 
and connect them by straight lines. What kind of a figure 
do these four lines enclose ? 

11. The side of a square — a ; the origin of co-ordinates 
is the intersection of the diagonals. What are the co-ordinates 
of the vertices (i.) if the axes are parallel to the sides of the 
square ? (li.) if the axes coincide with the diagonals ? 



C \ ( a a\ [ __2l a \ f— a — a \ ( a _ a V 
{1 ' } \2 2 J V 2 2/ V 2 2 J \2 2J ] 



(ii.)(^V2, 0), 0, ?V2, (--V2, 0], 0, -^V2 



Ans. 



2 J \ 2 J \ 2 

12. The side of an equilateral triangle = a ; the origin is 
taken at one vertex and the axis of x coincides with one side. 
What are the co-ordinates of the three vertices ? 



Ans. (0,0), (a, 0), f| |V5\ 



13. The line joining two points is bisected at the origin. 
If the co-ordinates of one of the points are a and b 7 what are 
the co-ordinates of the other ? 

14. Connect the points (5, 3) and (5, — 3) by a straight 
line. What is the direction of this line ? 



loci and their equations. 5 

Circular Measure. 

5. In Analytic Geometry, angles are often expressed in 
degrees, minutes, and seconds ; but sometimes it is very con- 
venient to employ the Circular Measure of an angle. 

In circular measure an angle is denned by the equation 

t arc 

angle = — - — > 
radius 

in which the word " arc " denotes the length of the arc corre- 
sponding to the angle when both arc and radius are expressed 
in terms of a common linear unit. 

This equation gives us a correct measure of angular magni- 
tude, because (as shown in Geometry) for a given angle the 
value of the above ratio of arc and radius is constant for all 
values of the radius. 

If the radius = 1, the equation becomes 

angle = arc ; that is, 

In circular measure an angle is measured by the length of 
the arc subtended by it in a unit circle. 

It is shown in Geometry that the circumference of a unit 
circle — 2 it ;. as this circumference contains 360° common 
measure, the two measures are easily compared by means 
of the relation 

360 degrees = 2 tt units, circular measure. 

Ex. 2. 

1. Find "the value in circular measure of the angles 1°, 
45°, 90°, 180°, 270°. 



Ans. — , J7r, 



180 



4 7 



, L 7T, 7T, 



2. In circular measure, unit angle is that angle whose arc 

is equal to the radius of the circle. "What is the value of this 

angle in degrees, etc. ? 

Ans. 57° 17' 45". 



ANALYTIC GEOMETRY. 



Distance between Two Points. 

6. To find the distance between two given points. 

Let P and Q (Fig. 2) be the given points, x x and y, the 
co-ordinates of P, x 2 and y 2 those of Q. Also let d = PQ = 
the required distance. 



r 


/ 


Q 




p/^ — 


R 









M 1 


\ r . 




Fig. 2. 

Traw PJ/and QX.L to OX, and PR II to OX 

Then OM = x u MP = y lf 

01f=x 2i NQ = y 2l 

PR = x 2 — x h QB = y 2 — y x . 

By Geometry, 



whence 



cZ 2 = (# 2 — 
d = V(as 2 



^i) 2 + (y 2 -yi) 2 ; 



• ^i) 2 + y, - 2/i) 2 - [1] 

Since (x x — # 2 ) 2 = : (#2 — #i) 2 , it makes no difference which 
point is called (x l9 y x ), and which (x 2 , y 2 ). 

7. Equation [1] is perfectly general, holding true for points 
situated in any quadrant. Thus, if P be in the second quad- 
rant and Q in the third quadrant (Fig. 3), x 2 —x 1 is obvi- 
ously equal to the leg RQ ; and since y 2 is negative, y 2 — y\ is 
the sum of two negative numbers, and is equal to the absolute 
length of the leg RP with the — sign prefixed. 



LOCI AND THEIR EQUATIONS. 7 

Note. The learner should satisfy himself that equation [1] is perfectly- 
general, by constructing other special cases in which the points Pand Q 
are in different quadrants. In every case he will find that the numeri- 
cal values of the expressions (x 2 ~ x i) an( l (Vt~ V\) are the legs of the 
right triangle the hypotenuse of which is the required distance PQ. 

Equation [1] is merely one illustration of a general truth, of which 
the learner will gradually become convinced as he proceeds with the 
study of the subject; namely, that theorems and formulas deduced by 
reasoning with points or lines in the first quadrant (where the co-ordinates 
are always positive) must, from the very nature of the analytic method, 
hold true when the points or lines are situated in the other quadrants. 

Ex. 3. 

Find the distance 

1. From the point (—2, 5) to the point (—8, —3). 

2. From the point (1, 3) to the point (6, 15). 

3. From the point (—4, 5) to the point (0, —2). 

4. From the origin to the point (—6, —8). 

5. From the point (a, b) to the point (—a, — 5). 

Find the lengths of the sides of a triangle 

6. If the vertices are the points (15,-4), (-9,3), (11,24). 

7. If the vertices are the points (2, 3), (4, —5), (-3, -6). 

8. If the vertices are the points (0, 0), (3, 4), (-3, 4). 

9. If the vertices are the points (0, 0), (—a, 0), (0, — b). 

10. The vertices of a quadrilateral are (5, 2), (3, 7), (—1, 4), 
(—3,-2). Find the lengths of the sides and also of the 
diagonals. 

11. One end of a line whose length is 13 is the point 
(—4, 8) ; the ordinate of the other end is 3. What is its 
abscissa ? 

12. What equation must the co-ordinates of the point (x, y) 
satisfy if its distance from the point (7, —2) is equal to 11 ? 



ANALYTIC GEOMETRY. 

13. What equation expresses algebraically the fact that 
le point (x } y) is equidistant from the points (2, 3) and 

:,5)? 

14. If the value of a quantity depends on the square of a 
.ngth, it is immaterial whether the length be considered 

)sitive or negative. Why ? 



Division of a Line. 

8. To bisect the line joining two given points. 

Let P and Q (Fig. 4) be the given points (x lf y Y ) and (x 2y y 2 ). 
9t x and y be the co-ordinates of P, the mid-point of PQ. 
The meaning of the problem is to find the values of x and y 
terms of x u y lt and x 2 , y 2 . 





S NX 



Fig. 4. 



Fig. 5. 



Draw PM, PS, QNJL to OX; also draw PA, PB || to OX. 
Then rt. A PPA = rt. A PQB (hypotenuse and one acute 
igle equal). 
Therefore PA = PB, and AR = BQ\ 

so - MS=SN. 

By substitution, x — x l — x 2 — x, and y — yi = y 2 — y\ 



aence 



[2] 



LOCI AND THEIR EQUATIONS. 9 

9. To divide the line joining two given points into two pai ts 
having a given ratio m : n. 

Let Pand Q (Fig. 5) be the given points (x\, y x ) and (x 2 , y 2 ). 
Let R be the required point, such that PR : RQ = m : n, and 
let x and y denote the co-ordinates of R. 

Complete the figure by drawing lines as in Fig. 4. 

The rt. A PR A and RQB, being mutually equiangular, 
are similar ; therefore 

PA^PR = m and AR^PR^m 
RB RQ n an BQ RQ~ n 

Substituting for the lines their values, we have 

- 1 — — , and £ ^ — — • 

x 2 — x n Vi~y n 

Solving these equations for x and y, we obtain 

rnoc 2 -f nx 1 my 2 + ny x rrtn 

If m = ?2, we have the special case of bisecting a line already 
considered ; and it is easy to see that the values of x and y 
reduce to the forms given in [2]. 

x. 4. 

What are the co-ordinates of the point 

1. Half-way between the points (5, 3) and (7, 9) ? 

2. Half-way between the points (-6, 2) and (4, -2) ? 

3. Half-way between the points (5, 0) and (—1, — 4) ? 

4. The vertices of a triangle are (2, 3), (4, -5), (-3,-6) ; 
find the middle points of its sides. 

5. The middle point of a line is (6, 4), and one end of the 
line is (5, 7). What are the co-ordinates of the other end ? 

6. A line is bisected at the origin ; one end of the line is the 
point (—a, b). What are the co-ordinates of the other end ? 



10 ANALYTIC GEOMETRY. 

7. Prove that the middle point of the hypotenuse of a right 
triangle is equidistant from the three vertices. 

8. Prove that the diagonals of a parallelogram mutually 
bisect each other. 

9. Show that the values of x and y in [2] hold true when 
the two given points both lie in the second quadrant. 

10. Solve the problem of § 9 when the line PQ is cut 
externally instead of internally, in the ratio m : n. 

11. What are the co-ordinates of the point which divides 
the line joining (3, —1) and (10, 6) in the ratio 3:4? 

12. The line joining (2, 3) and (4, —5) is trisected ; deter- 
mine the point of trisection nearest (2, 3). 

13. A line AB is produced to a point C, such that BC = 
\ AB. If A and B are the points (5, 6) and (7, 2), what are 
the co-ordinates of C? 

14. A line AB is produced to a point (7, such that AB : BC 
= 4:7. If A and B are the points (5,4) and (6, —9), what 
are the co-ordinates of C? 

15. Three vertices of a parallelogram are (1, 2), (—5, —3), 
(7, —6). "What is the fourth vertex ? 

Constants and Variables. 

10. In Analytic Geometry a line is regarded as a geometric 
magnitude traced or generated by a moving point, — just as we 
trace on paper what serves to represent a line to the eye by 
moving the point of a pen or pencil over the paper. 

We shall find that great advantages are to be gained by 
defining a line in this way, but we must be prepared from 
the outset to make an important distinction in the use of 
symbols representing lengths. We must distinguish between 
symbols which denote definite or fixed lengths and those 
which denote variable lengths. 



LOCI AND THEIR, EQUATIONS. 



11 



11. A simple example will serve to illustrate this difference. 
Let A (Fig. 6) be the point (3, 4). Then OA = V9 + 16 = 5. 
Now let a point P describe the line OA by moving from 
to A, and let the co-ordinates of P be denoted by x and y ; 
also let z denote the length OP at any position of P. Then 
it is clear that the distance OP or z will be equal to 0, to 
begin with, and will increase in value continuously until it 
becomes equal to 5. 




M B X 

Fig. 6. 



Here the word continuously deserves special attention. It 
means that P must pass successively through every conceiv- 
able position on the line OA from to A ; that, therefore, z 
must have in succession every conceivable value between 
and 5. There will be one position of P for which z is 
equal to 2 ; there will be another position of P for which 
z is equal to 2.000001 ; but before reaching this value it must 
first pass through all values between 2 and 2.000001. 

In the same way the co-ordinates of P, namely, x and y, 
both pass through a continuous series of changes in value 
unlimited in number, the abscissa x increasing continuously 
from to 3, and the ordinate y from to 4. 

We may now divide the lengths considered in this example 
into two classes : 

(1) Lengths supposed to remain constant in value, namely, 
the co-ordinates of J. and the distance OA ; (2) lengths sup- 
posed to vary continuously in value, namely, the co-ordinates 
of P, (x and y). and the distance OP or z. 



12 ANALYTIC GEOMETRY. 

Quantities of the first kind in any problem are called con- 
stant quantities, or, more briefly, Constants. 

Quantities of the second kind are called variable quantities, 
or, more briefly, Variables. 

12. Two variables are often so related that if one of them 
changes in value, the other also changes in value. The second 
variable is then said to be a function of the first variable. 
The second variable is also called the dependent variable, 
while the first is called the independent variable. Usually 
the relation between two variables is such that either may be 
treated as the independent variable, and the other as the 
dependent variable. 

Thus, in § 11, if we suppose z to change, then both x and 
y will change ; the values of x and y then will depend upon 
the value given to z ; that is, x and y will be functions of z. 
But we may also suppose the value of x, the abscissa of P, 
to change ; then it is clear that the values of both y and z 
must also change. In this case we take x as the independent 
variable, and values of y and z will depend upon the value 
of x ; that is, y and z will he functions of x. 

13. The most concise w T ay to express the relations of the 
constants and variables which enter into a problem is by 
means of algebraic equations. 

The co-ordinates of P (Fig. 6) throughout its motion are 
always x and y ; and the triangle OPM is similar to the 
triangle OAB. Hence, for any position of P, 

2 = -, and z 2 = x' + y 2 ] 
x 3 

whence, by solving, 

y = -x, and z = --x, 
J 3 4 

equations which express the values of y and z, respectively, 

in terms of x as the independent variable. 



LOCI AND THEIR EQUATIONS. 13 

14. In § 11, instead of assuming 3 and 4 as the co-ordi- 
nates of P, we might have employed two letters, as a and b, 
with the understanding that these letters should denote two 
co-ordinates wdiich remain constant in value during the motion 
of P. If we choose these letters, and then proceed exactly 
as in § 13, we obtain for the values of y and z, 

b Va 2 + b' 2 
y = -x. z = x. 

15. There is a noteworthy difference between the constants 
3 and 4 and the constants a and b. The numbers 3 and 4 
are unalterable in value ; they cannot be supposed to change 
under any circumstances. The letters a and b are constants 
in this sense only, that they do not change in value when Ave 
suppose x or y or z to change in value ; in other words, they 
are not functions of x or y or z in the particular problem 
under discussion. In all other respects they are free to rep- 
resent as many different values as we choose to assign to them. 

Constants of the first kind (arithmetical numbers) are called 
absolute constants. Constants of the second kind (letters) are 
called arbitrary or general constants. 

16. By general agreement, variables are represented by the 
last letters of the alphabet, as x, y, z\ while constants are 
most commonly represented by the first letters, a, b, c, etc., or 
by the last letters with subscripts added, as x x , y Y , x 2 , y 2 , etc. 

Ex. 5. 

1. A point P (x, y) revolves about the point Q (%. y{), keep- 
ing always at the distance a from it. Mention the constants 
and the variables in this case. What is the total change in 
the value of each variable ? 

2. A point Q (x, y) moves : first parallel to the axis of y, 
then parallel to the axis of x, then equally inclined to the 
axes. Point out in each case the constants and the variables. 



14 



ANALYTIC GEOMETRY. 



Locus of an Equation. 

17. Let us continue to regard x and y as the co-ordinates of 
a point, and proceed to illustrate the meaning of an algebraic 
equation containing one or both of these letters. 

Take as the first case the equation # — 4 = 0, whence # = 4. 
It is clear that this equation is satisfied by the co-ordinates of 
every point so situated that its abscissa is equal to 4 ; there- 
fore it is satisfied by the co-ordinate of every point in the line 



Fig. 7. 

AB (Fig. 7), drawn II to 07, on the right of OF, and at the 
distance 4 from OY. And it is also clear that this line con- 
tains all the points whose co-ordinates will satisfy the given 
equation. 

The line AB, then, may be regarded as the geometric rep- 
resentation or meaning of the equation x — 4 = 0; and con- 
versely, the equation x — 4 = may be considered to be the 
algebraic representative of this particular line. 

In Analytic Geometry the line AB is called the locus of 
the equation # — 4 = 0; conversely, the equation x — 4 = is 
known as the equation of the line AB. 

The line AB is to be regarded as extending indefinitely in 
both directions. If AB be described by a point P, moving 
parallel to the axis of y } then at all points x is constant in 



LOCI AND THEIR EQUATIONS. 



15 



value and equal to 4, while y (which does not appear in the 
given equation) is a variable, passing through an unlimited 
number of values, both positive and negative. 

18. The equation x — y = 0, or x = y, states in algebraic 
language that the abscissa of the point is always equal to the 
ordinate. 

By 



Values of x. 


Values of y. 


. . . 


. . 0. 


1 . . . 


. . 1. 


2 . . . 


. . 2. 


-1 . . . 


. . -1. 


etc. 


etc. 



Fig. 8. 



If we draw through the origin (Fig. 8) a straight line 
AB, bisecting the first and third quadrants, then it is easy to 
see that the given equation is satisfied by every 'point in this ■ 
line and by no other points. If we conceive a point P to 
move so that its abscissa shall always be equal to its ordinate, 
then the point must describe the line AB. In other words, 
if the point P is obliged to move so that its co-ordinates 
(which of course are variables) shall always satisfy the con- 
dition expressed by the equation x — y = ; then the motion 
of P is confined to the line AB. 

The line AB is the locus of the equation x — y = 0, and 
this equation represents the line AB. 

19. The equation 2x + y — 3 = is satisfied by an unlimited 
number of values of x and y. We may find as many of them 
as we please by assuming values for one of the variables, and 
computing the corresponding values of the other. 



16 



ANALYTIC GEOMETRY. 



If we assume for z the values given below, we easily find 
for y the corresponding values given in the next column. 




Values of x. Values of y. 

3. 

1 .... . 1. 

2 --1. 

3 -3. 

4 -5. 

-1 5. 

-2 7. 

-3 9. 

-4 11. 

etc. etc. 



Fig. 9. 



Plotting these points (as shown in Fig. 9), we obtain a series 
of points so placed that their co-ordinates all satisfy the given 
equation. By assuming for z values between and 1, 1 and 
2, etc., we might in the same way obtain as many points as 
we please between A and B, B and C, etc. In this case, 
however, the points all lie in a straight line (as will be shown 
later) ; so that if any two points are found, the straight line 
drawn through them will include all the points whose co-ordi- 
nates satisfy the given equation. Now imagine that a point JP, 
the co-ordinates of which are denoted by z and y, is required 
to move in such a way that the values of z and y shall always 
satisfy the equation 2z J r y — 3 — 0; then P must describe 
the line AB, and cannot describe any other line. 

The line AB is the locus of the equation 2z -\- y — 3 = 0. 

20. Thus far we have taken equations of the first degree. 
Let us now consider the equation z 2 — y 2 = 0. By solving for 
y, we obtain y = ± x. Hence for every value of x there are 



LOCI AND THEIR EQUATIONS. 



17 



two values of y, both equal numerically to x, but having unlike 
signs. Thus, for assumed values of x } we have corresponding 
values of y given below : 



Values of x. 


Values of y 


o, . . . 


. . 0. 


1 . . . 


. . i,-i. 


2 . . . 


2,-2. 


3 . . •. 


3,-3. 


-1. • • • 


. . -1, 1. 


-2 . . . 


. . -2, 2. 


-3 . . . 


. . -3, 3. 



zK 


T 


/b 




\0 


X 


A- 




c\ 



Fig. 10. 



By plotting a few points, and comparing this case with the 
example in § 19, it becomes evident that the locus of the- 
equation consists of two lines, AD, CD (Fig. 10), drawn 
through the origin so as to bisect the four quadrants. 

21. There is another way of looking at this case. The equa- 
tion x 2 — y 2 —0, by factoring, may be written (x—y) (%+y) = 0. 
Now the equation is satisfied if either factor = ; hence, it is 
satisfied if x — y = 0, and also if x -f y = 0. We know (see 
§ 19) that the locus of the equation x — y = is the line 
AD (Fig. 8). And the locus of the equation x -J- y = (or 
x — ~y) is evidently the line CD, since every point in it is 
so placed that the two co-ordinates are equal numerically but 
unlike in sign. Therefore the original equation x 2 — y 2 = 
is represented by the pair of lines AD and CD (Fig. 10). 



22. Let us next consider the equation x 2 -f y 
ing for y, we obtain y = =b V25 — x 



25. Solv- 
When x < 5 there 



are two values of y equal numerically but unlike in sign. 
When x = 5, y = 0. When x > 5 the values of y are imagi- 
nary ; this last result means that there is no point with an 
abscissa greater than 5 whose co-ordinates will satisfy the 
given equation. 



18 



ANALYTIC GEOMETRY. 



By assigning values of x differing by unity, we obtain the 
following sets of values of x and y ; and by plotting the points, 
and then drawing through them a continuous curve, we obtain 
the curve shown in Fig. 11. 




Values of x. 


Values of y. 


o . . . 


. . ±5. 


l . . . 


. . ± 4.9. - 


2 . . . 


. . ± 4.6. 


3 . . : 


. . ±4. 


4 . . . 


. . ±3. 


5 . . . 


. . 0.- 


-1 . . . 


. . ±4.9. 


— 2 . . 


. . ± 4.6. 


-3 . . 


. . ±4. 


-4 . . 


. . ±3. 


-5 . . 


. . 0. 



Fig. II. 



In this case, however, the locus may be found as follows : 
Let P (Fig. 11) be any point so placed that its co-ordinates, 
x = OM, y = MP, satisfy the equation x 2 + y 2 = 25. Join 
OP; then x 2 + y 2 = OP 2 ; therefore OP =5. Hence, if P is 
anywhere in the circle described with as centre and 5 for 
radius, its co-ordinates will satisfy the given equation ; and 
if P is not in this circle, its co-ordinates will not satisfy the 
equation. This circle, then, is the locus of the equation. 

23. The points whose co-ordinates satisfy the equation 
y 2 = 4:X lie neither in a straight line nor in a circle. Never- 
theless, they do all lie in a certain line, which is, therefore, 
completely determined by the equation. To construct this 
line, we first find a number of points which satisfy the equa- 
tion (the closer the points to one another, the better) and then 
draw, freehand or with the aid of tracing curves, a continuous 
curve through the points. 

The co-ordinates of a number of such points are given in 



LOCI AND THEIJR EQUATIONS. 



19 



the table below. It is evident that for each positive value of 
x there are two values of y, equal numerically but unlike in 
sign. If we assume a negative value for z, then the value 
of y is imaginary ; this result means that there are no 
points to the left of the axis of y which will satisfy the given 
equation. 

Values of x. Values of y. 

0. 

1 ±2. 

2 ±2.83. 

3 ±3.46. 



4 
5 
6 
7 
8 
9 
1 



±4. 
± 4.47. 
± 4.90. 
± 5.29. 
± 5.66. 
±6. 
imaginary. 




Fig. 12. 



In Fig. 12 the several points obtained are plotted, and a 
smooth curve is then drawn through them. It passes through 
the origin, is placed symmetrically on both sides of the axis 
of x, lies wholly on the right of the axis of y, and extends 
towards the right without limit. It is the locus of the given 
equation, and is a curve called the Parabola. 

24. After a study of the foregoing examples, we may lay 
down the following general principles, which form the foun- 
dation of the science of Analytic Geometry : 

I. Every algebraic equation involving x and y is satisfied 
by an unlimited number of sets of values of x and y ; in other 
words, x and y may be treated as variables, or quantities vary- 
ing continuously, yet always so related that their values con- 
stantly satisfy the equation. 



20 ANALYTIC GEOMETRY. 

II. The letters x and y may also be regarded as represent- 
ing the co-ordinates of a point. This point is not fixed in 
position, because x and y are variables ; but it cannot be 
placed at random, because x and y can have only such values 
as will satisfy the equation ; now since these values are con- 
tinuous, the point may be conceived to move continuously, and 
will therefore describe a definite line, or group of lines, 

III. The line, or group of lines, described by a point 
moving so that its co-ordinates always satisfy the equation is 
called the Locus of the Equation ; conversely, the equation sat- 
isfied by the co-ordinates of every point in a certain line is 
called the Equation of the Line. 

IV. An equation, therefore, containing the variables x and 
y is the algebraic representation of a line ; it determines a 
certain line in the same sense that two co-ordinates determine 
a certain point. And, conversely, the line (or group of lines) 
which is the locus of an equation is the geometric representa- 
tion of the equation. 

Ex. 6. 

Determine and construct the loci of the following equations 
(the locus in each case being either a straight line or a circle) : 

1. x- 6 = 0. 9. 9^ 2 -25-0. 

2. 2; + 5 = 0. 10. Ax 2 -y 2 = 0. 

3. y = -l. 11. ^ 2 — 16y 2 = 0. 

4. x = 0. 12. x 2 + y 2 =^5>6. 

5. y = 0. 13. x 2 + y 2 -l = 0. 

6. x + y = 0. 14. x(y+5) = 0. 

7. x-2y = 0. 15. (x-2J(x— 3) = 0. 

8. 2^ + 3y+10 = 0. 16. ( y _4) (y + 1) = 0. 



LOCI AND THEIR EQUATIONS. 21 

17. What is the geometric meaning of the equation 
5a; 2 -17a;-12 = 0? 

Hint. Resolve the equation into two binomial factors. 

18. What is the geometric meaning of the equation 

19. What two lines form the locus of the equation 
xy+±x = 0? 

20. Is the point (2, — 5) situated in the locus of the equa- 
tion 4 a; - 3y - 22 = ? 

Hint. See if the co-ordinates of the point satisfy the equation. 

21 . Is the point (4,-6) in the locus of the equation y 2 = 9 x ? 

22. Is the point (— 1, —1) in the locus of the equation 
16x 2 + 9y 2 + 15a; — 6y -18 = 0? 

23. Does the locus of the equation x 2 -\-y 2 —100 pass through 
the point (—6,8)? 

24. Which of the loci represented by the following equa- 
tions pass through the origin ? 

(1) 3#+2 = 0. (5) 3a; = 2y. 

(2) 3a;-lly + 7 = 0. (6) 3x — lly = 0. 

(3) a; 2 -16y 2 -10 = 0. (7) ar -16?/ 2 = 0. 

(4) ax + by + c=--0. . (8) ax + by = 0. 

25. The abscissa of a point in the locus of the equation 
3 a; — 4:y — 7 = is 9; what is the value of the ordinate ? 

Arts. 5. 

26. Determine that point in the locus of y 2 — 4a; =0 for 

w T hich the ordinate == — 6. A m U . . /n CN 

Arts, the point (9, — 6). 

27. Determine the point where the line represented by the 
equation 7x + y — 14 = cuts the axis of x. 

Ans. The point (2, 0). 



22 analytic geometry. 

Intersections of Loci. 

25. The term Curve, as used in Analytic Geometry, means 
any geometric locus, including the straight line as well as 
lines commonly called curves. 

The Intercepts of a curve on the axes are the distances from 
the origin, measured along the axes, to the points where the 
curve cuts the axes. 

23. To find the intercepts of a curve, having given its equation. 

The intercept of a curve on the axis of x is the abscissa of 
the point where the curve cuts the axis of x. The ordinate of 
this point == 0. Therefore, to find this intercept, put y — in 
the given equation of the curve, and then solve the equation 
for x ; the resulting value of x will be the intercept required. 

If the equation is of a higher degree than the first there 
will be more than one value of x ; and the curve will cut the 
axis of x in as many points as there are real values of x. 

If an imaginary value of x is found it is to be rejected. 
But in order to make the language of geometry correspond to 
that of algebra, we may say in this case that the curve cuts 
the axis of x in an imaginary point ; that is, a point that has 
one or both of its co-ordinates imaginary. 

Similarly, to find the intercepts on the axis of y, put #=0 
in the given equation, and then solve it for y ; the resulting 
real values of y will be the intercepts required. 

27. To find the points of intersection of two curves, having 
given their equations. 

Since the points of intersection lie in both curves, their co- 
ordinates must satisfy both equations. Therefore, to find their 
co-ordinates, solve the two equations, regarding the variables 
x and y as unknown quantities. 

If the equations are both of the first degree, there will be . 



LOCI AND THEIR EQUATIONS. 23 

only one pair of values of x and y, and one point of inter- 
section. 

If the equations are, one or both of them, of higher degree 
than the first, there may be several pairs of values of x and 
y ; in this case there will be as many points of intersection as 
there are pairs of real values of x and y. 

If imaginary values of either x or y are obtained, there are 
no corresponding points of intersection. 

28. If a curve pass through the origin, its equation, reduced 
to its simplest form, cannot have a constant term ; that is, cannot 
have a term free from both x and y. 

Since in this case the point (0, 0) is a point of the curve, 
its equation must be satisfied by the values x = 0, and y = 0. 
But it is obvious that these values cannot satisfy the equation 
if, after reduction to its simplest form, it still contains a con- 
stant term. Therefore the equation cannot have a constant 
term. 

29. Jf an equation has no constant term, its locus must pass 
through the origin. 

For, the values x — 0, y = must evidently satisfy the equa- 
tion, and therefore the point (0, 0) must be a point of the 
locus. 

Ex. 7. 

Find the intercepts of the following curves : 

1. 4^+3y-48 = 0. 

2. by- Sx -30-0. 

3. x 2 ~\- y 2 = 16. 

4. 9x 2 + 4:y 2 =l6. 

5. 9x 2 -iy 2 = 16. 12. x 2 + if- 4#- 8y -= 32. 

6. 9x 2 -4y = 16. 13. x 2 + y 2 - 4#- 8y = 0. 

7. a 2 x 2 + by = a 2 b 2 . 14. (x - 5) 2 + (y— 6) 2 = 20. 



8. 


x - 3 = 0. 


9. 


x* -9 = 0. 


10. 


x*-tf = 0. 


11. 


if = 4x. 


12. 


X 2 + 7/-4:i 



24 ANALYTIC GEOMETRY. 

Find the points of intersection of the following curves : 

15. 3a?-4y + 13 = 0, 11a; + 7y — 104 = 0. 

16. 2a; + 3y = 7, x — y==l. 

17. a;-7y+25 = 0, x 2 + y 2 = 25. 

18. 3a; + 4y = 25, a; 2 + y 2 = 25. 

19. a; + y = 8, x 2 ^-y 2 = M. 

20. 2x = y, x 2 + ?f-10x = 0. 

21. The equations of the sides of a triangle are 2a; + 9y 
+ 17 = 0, 7 a?- y- 38 = 0, x— 2y + 2 = 0. Find the co- 
ordinates of its three vertices. 

22. The equations of the sides of a triangle are 5 a; + 6 y = 12, 
3a; — 4y = 30, x + 5y = 10. Find the lengths of its sides. 

23. Find the lengths of the sides of a triangle if the equa- 
tions of the sides are x = 0, y = 0, and 4a; -f- 3y = 12. 

24. What are the vertices of the quadrilateral enclosed by 
the straight lines x — a = 0, a? + a = 0, y — 6 = 0, y + 6 = 0? 
"What kind of a quadrilateral is it? 

25. Does the straight line 5a;-f4y = 20 cut the circle 
a; 2 + y 2 = 9? 

26. Find the length of that part of the straight line 
3a; — 4y =. which is contained within the circle x 2 -{-y 2 = 25. 

27. Which of the following curves pass through the origin 
of co-ordinates ? 

(1) 7a;-2y+4 = 0. (4) ax + by = 0. 

(2) 7 a; — 2y = 0. (5) ax + by + c = 0. 

(3) y 2 -a; 2 = 4y. (6) x 2 — y + a = a + xy. 

28. Change the equation 4a; + 2y — 7 = so that its locus 
shall pass through the origin. 



loci and their equations. 25 

Construction of Loci. 

30. If we know that the locus of a given equation is a 
straight line, the locus is easily constructed ; it is only neces- 
sary to find any two points in it, plot them, and draw a 
straight line through them with the aid of a ruler. 

Likewise, if we know that the locus is a circle, and can find 
its centre and its radius, the entire locus can then be. immedi- 
ately described with the aid of a pair of compasses. 

It will appear later on that the form of the given equation 
enables us at once to tell whether its locus is a straight line 
or a circle. 

If the locus of an equation is neither a straight line nor a 
circle, then the following method of construction, which is 
applicable to the locus of any equation without regard to the 
form of the curve, is usually employed. 

31. To construct the locus of a given equation. 

The steps of the process are as follows : 

1. Solve the equation with respect to either x or y. 

2. Assign values to the other variable, differing not much 
from one another. 

3. Find each corresponding value of the first variable. 

4. Draw two axes, choose a suitable scale of lengths, and 
plot the points whose co-ordinates have been obtained. 

5. Draw a continuous curve through these points. 

Discussion. An examination of the equation, as shown in 
the examples given below, enables us to obtain a good general 
idea of the shape and size of the curve, its position with respect 
to the axes, etc. ; in this way it serves as an aid in construct- 
ing the curve, and as a means of detecting numerical errors 
made in computing the co-ordinates of the points. Such an 
examination is called a discussion of the equation. 



26 



ANALYTIC GEOMETRY. 



Note 1. This method of constructing a locus is from its nature an 
approximate method. But the nearer the points are to one another, the 
nearer the curve will approach the exact position of the locus. 

Note 2. In theory, it is immaterial what scale of lengths is used. 
In practice, the unit of lengths should be determined by the size of the 
paper compared with the greatest length to be laid off upon it. Paper 
sold under the name of " co-ordinate paper," ruled in small squares, y 1 ^ 
of an inch long, on each side, will be found very convenient in practice. 

32. Construct the locus of the equation 
9x 2 + 4y 2 - 576 = 0. 
If we solve for both x and y, we obtain the following values : 

(1) 
(2) 



y = -j. f V64 _ x l , 



# = =b f Vl44 - y 2 . 

By assigning to x values differing by unity, and finding 
corresponding values of y, we obtain the results given below. 
To each value of x, positive or negative, there correspond two 
values of y, equal numerically and unlike in sign. By plotting 
the corresponding points, and drawing a continuous curve 
through them, we obtain the closed curve shown in Fig. 13. 




lnes of x. 




Values of y. 


... ±12. 


±1 






±11.91. 


±2 






± 11.62. 


d=3 






± 11.13. 


±4 






± 10.39. 


±5 






± 9.36. 


±6 






± 7.93. 


±7 






± 5.80. 


±8 






± 0. 


±9 






± imaginary 



LOCI AND THEIR EQUATIONS. 27 

Discussion. From equations (1) and (2) we see that if 
x = 0, y = zb 12, and if y — 0, # = zb 8 ; therefore the inter- 
cepts of the curve on the axis of x are + 8 and — 8, and those 
on the axis of y are +12 and —12. These intercepts are the 
lengths OA, OA\ and OB, OB 1 , in Fig. 13. 

If we assign to x a numerical value greater than 8, posi- 
tive or negative, we find by substitution in equation (1) that 
the corresponding value of y will be imaginary. This shows 
that OA and OA' are the maximum abscissas of the curve. 
Similarly, equation (2) shows that the curve has no points 
with ordinates greater than +12 and —12. 

The greater the numerical value of x, between the limits 
and + 8 or and —8, the less the corresponding value of y 
numerically ; why ? 

From equation (1) we see that corresponding to each 
value of x, between the limits and ± 8, there are two real 
values of y, equal numerically and unlike in sign. Hence, for 
each value of x between and db 8 there are two points of the 
curve placed equally distant from the axis of x. Therefore 
the curve is symmetrical with respect to the axis of x ; in 
other words, if the portion of the curve above the axis of x 
be revolved about this axis through 180°, it will coincide with 
the portion below the axis. Similarly, it follows from equa- 
tion (2) that the curve is also symmetrical with respect to 
the axis of y. Therefore the entire curve is a closed curve, 
consisting of four equal quadrantal arcs symmetrically placed 
about the origin 0. The name of this curve is the Ellipse. 

33. Construct the locus of the equation 
4.r-y 2 + 16 = 0. 

Solving for both x and y, we obtain 

y = =b2V^+4, (1) 

.-i=3t (2 ) 



ANALYTIC GEOMETRY. 



We may either assign values to x, and then compute those 
of y by means of (1), or assign values to y, and compute those 
of x by means of (2) ; the second course is better, because 
there is less labor in squaring a number than in extracting 
its square root. 

By assigning values to y, differing by unity from to +10, 
and from to —10, and then proceeding exactly as in the 
last example, we obtain the series of values given below, and 
the curve shown in Fig. 14. 




Fig. 14. 



ues of y. 


Values of x. 


o . . . 


. . -4. 


±i . . . 


. . -3.75. 


±2 . . . 


. . -3. 


±3 . . . 


. . -1.75. 


±4 . . . 


. . 0. 


±5 . . . 


. . 2.25. 


±6 . . . 


. . 5. 


±7 . . . 


. . 8.25. 


±8 . . . 


. . 12. 


±9 . . . 


. . 16.25. 


= 10 . . . 


. . 21. 



Discussion. An examination of equations (1) and (2) 
yields the following results, the reasons for which are left 
as an exercise for the learner : 

The intercepts on the axes are : 

On the axis of x, OA = — 4. 

On the axis of y, OD = + 4, and OC= - 4. 

If we draw through A the line AD _L to OX, the entire 
curve lies to the right of AD. 

The curve is situated on both sides of OX, and is sym- 
metrical with respect to OX. 

The curve extends towards the right without limit. 



LOCI AND THEIR EQUATIONS. 



29 



The curve constantly recedes from OX as it extends 
towards the right. 

This curve is called a Parabola; the point A is called its 
Vertex, the line AX\\& Axis. 

34. Construct the locus of the equation 
y — sin x. 

If we assume for x the values 0°, 10°, 20°, 30°, etc., the cor- 
responding values of y are the natural^s'mes of these angles, 
and are as follows : 



Values of x. 


Values of y. 


Values of x. 


Values of y 


0° .'. . 


. . 0. 


50° . . 


. . 0.77. 


10° . . 


. . 0.17. 


60° . . 


. . 0.87. 


20° . . 


. . 0.34. 


70° . . 


. . 0.94. 


30° . . 


. . 0.50. 


80° . . 


. . 0.98. 


40° . . 


. . 0.64. 


90° . . 


. . 1. 



If we continue the values of x from 90° to 180°, the above 
values of y repeat themselves in the inverse order (e.g., if 
x = 100°, y = 0.98, etc.) ; from 180° to 360° the values of y 
are numerically the same, and occur in the same order as 
between 0° and 180°, but are negative. 




Fig. 15. 



In order to express both x and y in terms of a common 
linear unit, we ought, in strictness, to use the circular meas- 
ure of an angle in which the linear unit represents an angle 



30 ANALYTIC GEOMETRY. 

of 57.3°, very nearly (see § 5). But it is more convenient, 
and serves our present purpose equally well, to assume that 
an angle of 60° — the linear unit. This assumption is made 
in Fig. 15, where the curve is drawn with one centimeter as 
the linear unit. 

Discussion. The curve passes through the origin, and cuts 
the axis of x at points separated by intervals of 180°. Since 
an angle may have any magnitude, positive or negative, the 
curve extends on both sides of the origin without limit. The 
maximum value of the ordinate is alternately +1 and —1 : 
the former value corresponds to the angle 90°, and repeats 
itself at intervals of 360° ; the latter value corresponds to 
the angle 27,0°, and repeats itself at intervals of 360°. The 
curve has the form of a wave, and is called the Sinusoid. 

Ex. 8. 

Construct the loci of the following equations : 

1. 3^-y-2 = 0. 13. 3/ 2 -l = 0. 

2. y = 2x. 14. y = x*. 

3. x 2 = y 2 . 15. an/ = 12. 

4. x 2 + y 2 = 100. 16. x — sin y. 

5. x 2 — y 2 = 25. 17. y = 2 sin x. 

6. 4^ — ^ = 0. 18. y = sin 2x. 

7. 4^ + 9?/ 2 = 144. 19. y = cosx. 

8. y 2 -\6x = 0. 20. y = tsmx. 

9. y 2 + 16x^=0. 21. y = cotx. 

10. x 2 — 2x—10y—5 = 0. 22. y = secx. 

11. zj 2 — 2y-10x = 0. 23. y = cscx. 

12. (#-3) 2 +(y-2) 2 = 25. 24. y = sinx + cosx. 



loci and their equations. 31 

Equation of a Curve. 

35. From what precedes, we may conclude that every equa- 
tion involving x and y as variables represents a definite line 
(or group of lines) known as the locus of the equation. 
Regarded from this point of view, an equation is the state- 
ment in algebraic language of a geometric condition w T hich 
must always be satisfied by a point (x, y), as we imagine it 
to move in the plane of the axes. For example, the equation 
x = 2y states the condition that the point must so move that 
its abscissa shall always be equal to twice its ordinate ; the 
equation x l + y 2 = 4 states the condition that the point must 
so move that the sum of the squares of its co-ordinates shall 
always be equal to 4 ; etc. 

Conversely, every geometric condition that a point is 
required to satisfy must confine the point to a definite line 
as its locus, and must lead to an equation that is always 
satisfied by the co-ordinates of the point. 

Hence arises a new problem, and one usually of greater 
difficulty than any thus far considered, namely : 

Given the geometric condition to he satisfied by a point, to 
find the equation of its locus. 

The great importance of this problem lies in the fact that 
in the practical applications of Analytic Geometry the law of 
a moving point is commonly the one thing known to start 
with, so that the first step must consist in finding the equa- 
tion of its locus. 

Ex. 9. 

1. A point moves so that it is always three times as far from 
the axis of x as from the axis of y. What is the equation of 
its locus? 

2. What is the equation of the locus of a point which moves 
so that its abscissa is always equal to + 6 ? — 6? 0? 



32 ANALYTIC GEOMETRY. 

3. What is the equation of the locus of a point which moves 
so that its ordinate is always equal to +4? — 1? 0? 

4. A point so moves that its distance from the straight line 
x — 3 is always numerically equal to 2. "What is the equation 
of its locus ? 

5. A point so moves that its distance from the straight line 
y =■ 5 is always numerically equal to 3. Find the equation 
of its locus. Construct the locus. 

6. A point moves so that its distance from the straight 
line x + 4 = is always numerically equal to 5. Find the 
equation of its locus. Construct the locus. 

7. What is the equation of the locus of a point equidistant 

(1) from the parallels x = and x ■= — 6 ? 

(2) from the parallels y = 7 and y = — 3 ? 

8. What is the equation of the locus of a point always 
equidistant from the origin and the point (6, 0) ? 

Find the equation of the locus of a point 
9. Equidistant from the points (4, 0) and (— 2, 0). 

10. Equidistant from the points (0, — 5) and (0, 9). 

11. Equidistant from the points (3, 4) and (5, —2). 

12. Equidistant from the points (5, 0) and (0, 5). 

13. A point moves so that its distance from the origin is 
always equal to 10. Find the equation of its locus. 

14. A point moves so that its distance from the point (4,-3) 
is always equal to 5. Find the equation of its locus, and con- 
struct it. What kind of curve is it ? Does it pass through 
the origin ? Why ? 

15. What is the equation of the locus of a point whose 
distance from the point (— 4, — 7) is always equal to 8 ? 



LOCI AND THEIR EQUATIONS. 33 

16. About the origin of co-ordinates as centre, with a radius 
equal to 5, a circle is described. A point outside this circle 
so moves that its distance from the circumference of the circle 
is always equal to 4. What is the equation of its locus ? 

17. A high rock A, rising out of the water, is 3 miles 
from a perfectly straight shore BC. A vessel so moves that 
its distance from the rock is always the same as its distance 
from the shore. What is the equation of its locus ? 

18. A point A is situated at the distance 6 from the line 
BC. A moving point P is always equidistant from A and 
BC. Find the equation of its locus. 

19. A point moves so that its distance from the axis of x is 
half its distance from the origin ; find the equation of its locus. 

20. A point moves so that the sum of the squares of its 
distances from the two fixed points (a, 0) and (— a, 0) is the 
constant 2 k 2 ; find the equation of its locus. 

21. A point moves so that the difference of the squares of 
its distances from (a, 0) and (— a, 0) is the constant k 2 ; find 
the equation of its locus. 

Ex. 10. (Review.) 

1. If we should plot all possible points for w T hich x = — 5, 
how would they be situated ? 

2. Construct the point (x, y) if x = 2 and 

(1) y = 4#-3, (2) 3^-2y-8. 

3. The vertices of a rectangle are the points (a, b), (—a, b), 
(—a, —6), and (a, —b). Find the lengths of its sides, the 
lengths of its diagonals, and show that the vertices are equi- 
distant from the origin. 

4. What does equation [1], p. 6, for the distance between 
two points, become when one of the points is the origin ? 



34 ANALYTIC GEOMETRY. 

5. Express by an equation that the distance of the point 
(x, y) from the point (4, 6) is equal to 8. 

6. Express that the point (x, y) is equidistant from the 
points (2, 3) and (4, 5). 

7. Find the point equidistant from the points (2, 3), (4, 5), 
and (6, 1). What is the common distance ? 

8. Prove that the diagonals of a rectangle are equal. 

9. Prove that the diagonals of a parallelogram mutually 
bisect each other. 

10. The co-ordinates of three vertices of a parallelogram 
are known : (5, 3), (7, 10), (13, 9). "What are the co-ordi- 
nates of the remaining vertex ? 

11. The co-ordinates of the vertices of a triangle are (3, 5), 
(7, —9), (2, —4). Find the co-ordinates of the middle points 
of its sides. 

12. The centre of gravity of a triangle is situated on the 
line joining any vertex to the middle point of the opposite 
side, at the point of trisection nearest that side. Find the 
centre of gravity of the triangle whose vertices are the points 
(2, 3), (4,-5), (-3, -6). 

13. The vertices of a triangle are (5, - 3), (7, 9), (- 9, 6). 
Find the distance from its centre of gravity to the origin. 

14. The vertices of a quadrilateral are (0, 0), (5, 0), (9, 11), 
(0, 3). Find the co-ordinates of the intersection of the two 
straight lines which join the middle points of its opposite sides. 

15. Prove that the two straight lines which join the middle 
points of the opposite sides of any quadrilateral mutually 
bisect each other. 

16. A line is divided into three equal parts. One end of 
the line is the point (3, 8) ; the adjacent point of division is 
(4, 13). What are the co-ordinates of the other end? 



LOCI AND THEIR EQUATIONS. 35 

17. The line joining the points (x Xl y x ) and (x 2 , y 2 ) is divided 
into four equal parts. Find the co-ordinates of the points of 
division. 

18. Explain and illustrate the relation which exists between 
an equation and its locus. 

19. Construct the two lines which form the locus of the 
equation x 2 — 7x = 0. 

20. Is the point (2, — 5) in the locus of the equation 
4^ 2 -9y 2 =:36? 

21. The ordinate of a certain point in the locus of the 
equation x 2 + y 2 + 20x — 70 = is 1. What is the abscissa 
of this point ? 

22. Find the intercepts of the curve x 2 -\-y 2 — hx— 7y + 6 = 0. 

Find the points common to the curves : 

23. x 2 + y 2 = \00 } and y 2 — 12a; = 0. 

24. x 2 + y 2 = 5 a 2 , and x 2 = 4 ay. 

25. b 2 x 2 -f- a 2 y 2 == d 2 b 2 , and x 2 -f y 2 = o?> 

26. Find the lengths of the sides of a triangle, if its ver- 
tices are (6,0), (0, -8), (-4, -2). 

27. A point moves so that it is always six times as far from 
one of two fixed perpendicular lines as from the other. Find 
the equation of its locus. 

28. A point so moves that its distance from the fixed 
point A is always double its distance from the fixed line AB. 
Find the equation of its locus. 

29. A fixed point is at the distance a from a fixed straight 
line. A point so moves that its distance from the fixed point 
is always twice its distance from the fixed line. Find the 
equation of its locus. 



CHAPTER II. 

THE STRAIGHT LINE. 
Equations of the Steaight Like. 

36. Notation. Throughout this chapter, and generally in 
equations of straight lines, 

a = the intercept on the axis of x. 

b — the intercept on the axis of y. 

y — the angle between the line and the axis of x. 
m = tan y. 
p = the distance of the line from the origin. 

a = the angle between p and the axis of x. 

These six quantities are constants for a given straight line, 
but vary in value for different lines : a, b, m, and p may 
have any values from — oo to -fa> ; y and a, any values from 
0° to 360°. 

The constant m is often called the Slope of the line ; its 
value determines the direction of the line. 

In order to determine a straight line, two geometric con- 
ditions must be given. 

37. To find the equation of a straight line passing through 
tivo given points (x^ y x ) and (x 2 , y 2 ). 

Let A (Fig. 16) be the point (x lt y Y ), B the point (x 2l y 2 ) ; 
and let P be any other point of the line drawn through A 
and B, x and y its co-ordinates. Draw A C, BB } PM, J- to 
OX, andAS^IIto OX 



THE STRAIGHT LINE. 



37 



The triangles APF, ABE are similar ; therefore 

PF_ = BE 
AF AE 

Now PF=y — y,, AF=x—x 1 , BE=y. i —y l , AE—x^—x^. 
Therefore l^g-g^ [4] 

This is the equation required. 




F 
sA E 


_F 


C $ 




o j x 



Fig, 16. 



Fig. 17. 



It is evident that the angle PAF=y. Therefore each 
side of equation [4] is equal to tan y or m. The first side 
contains the two variables x and y, and the equation tells us 

that they must vary in such a way that the fraction ^ ^ 

X X\ 

shall remain constant in value, and always equal to m. 

Note. In Fig. 16 the points A, B, and P are assumed in the first 
quadrant in order to avoid negative quantities. But the reasoning will 
lead to equation [4] whatever be the positions of these points. In 
Fig. 17 the points are in different quadrants. The triangles APF, 
ABE are to be constructed as shown in the figure. They are 
similar ; and by taking proper account of the algebraic signs of the 
quantities, we arrive at equation [4], as before. The learner should 
study this case with care, and should study other cases devised by him- 
self, till he is convinced that equation [4] is perfectly general. 



38 



ANALYTIC GEOMETRY. 



38. To find the equation of a straight line, given one point 
(x ly y x ) in the line and the angle y. 

Let the figure be constructed like Fig. 16, omitting the 
point B and the line BED. Then it is evident that 



m 



-PF _ y-y x . 



whence, 




AF x — xi 
V — Vi = m(x — x 1 ). 



B 



[5] 







Fig. 18. 



M A\ X 



Fig. 19. 



39. To find the equation of a straight line, given the intercept 
b and the angle y. 

Let the line cut the axes in the points A and B (Fig. 18). 
Let P be any point (#, y) in the line. Draw PM± to OX, 
and BO il to OX. 

Then OB = b, PBC=y, BC=x, PC=y-b', 

therefore m = ; 

x 

whence V = mx + b. [6] 

40. To find the equation of a straight line, given its intercepts 
a and b. 

Let the line cut the axes in the points A and B (Fig. 19), 
and let P be any other point (x, y) in the line. Then OA = a, 



THE STRAIGHT LINE. 



39 



OB = b. Draw PM 1. to OX. The triangles PMA, BOA 
are similar ; therefore 

PM MA OA - OM 



BO OA 



OA 



or 



y 




a — 


# 


b 




a 




as 

a 


+ 


b " 


= 1. 



1-- 



whence — + 7 ; = 1. [7] 

This is called the Symmetrical Equation of the straight line. 

41. To find the equation of a straight line, given its distance 
p from the origin, and the angle a. 



F 


B 


s>S 






/' 




Nv4 







M 


\x 



Fig. 20. 

Let AB (Fig. 20) be the line, P any point in it. Draw 
OS -L to AB, meeting AB in S\ PM ± to OX; MP II to 
AB, meeting OS in P ; and PQ J. to AB. 

Then jp = OS = OP + QP, a = 80M= PMQ. 

By Trigonometry 

OP = OM cos a = x cos a, 
QP =?= PJ/sin a = y sin a. 
Therefore OP + QP == ;? = # cos a + y sin a. 

Or » cos a + y sin a = p m [8] 

This is called the Normal Equation of the straight line. 

The quantity p is always positive (like the radius of unit 
circle in Trigonometry). 



v 



40 ANALYTIC GEOMETRY. 

Note 1. Observe that all the equations of the straight line which 
have been obtained are of the first degree. 

Note 2. For the value of the sines, cosines, and tangents of the more 
common angles, see Appendix. 

Ex. 11. 

Find the equation of the straight line passing through the 
two points : 

1. (2, 3) and (4, 5). 7. (2, 5) and (0, 7). 

2. (4, 5) and (7, 11). 8. (3, 4) and (0, 0). 

3. (-1,2) and (3, -2). 9. (3, 0) and (0, 0). 

4. (-2, -2) and (-3,-3). 10. (3, 4) and (-2, 4). 

5. (4, 0) and (2, 3). 11. (2, 5) and (- 2, - 5). 

6. (0, 2) and (-3, 0). 12. (m, n) and (-m, — n). 

Find the equation of a straight line, given : 



13. 


(4, l)andy = 45°. 


29. 


5 = -4, y = 120°. 


14. 


(2, 7) andy = 60°. 


30. 


b = — 4, y=135°. 


15. 


(-3, 11) and y = 45°. 


81. 


6 = -4, y=150°. 


16. 


(13,-4) and y = 150°. 


32. 


& = — 4, y = 225°. 


17. 


(3, 0)andy = 30°. 


33. 


a = 4, 6 = 3. 


18. 


(0, 3) and y= 135°. 


34. 


a = -6, 6 = 2. 


19. 


(0, 0) and y = 120°. 


35. 


a = -3, 6 = -3. 


20. 


(2,-3) andy = 0°. 


36. 


a = 5, 6 = — 3. 


21. 


(2, - 3) and y = 90°. 


37. 


a = — 10, 6 = 5. 


22. 


6 = 2andy = 45°. 


38. 


o=l, 6 = -l. 


23. 


5 = 5, y=45°. 


39. 


a = n, 6 = — n. 


24. 


6 = -4, y = 45°. 


40. 


a = n, b = 4n. 


25. 


6 = -4, y=30°. 


41. 


p = 5, a = 45°. 


26. 


6 = -4,y = 0°. 


42. 


p = 5, a =120°. 


27. 


6 = -4, y = 60°. 


43. 


p = 5, a = 240°. 


28. 


5 = -4, y = 90°. 


44. 


p = 5, a = 300°. 



THE STRAIGHT LINE. 41 

Write the equations of the sides of a triangle : 

45. If its vertices are the points (2, 1), (3, — 2), (-4, -1). 

46. If its vertices are the points (2, 3), (4, — 5), (— 3, — 6). 

47. Form the equations of the medians of the triangle 
described in No. 53. 

48. The vertices of a quadrilateral are (0, 0), (1, 5), (7, 0), 
(4, — 9). Form the equations of its sides, and also of its 
diagonals. 

Find the equation of a straight line, given : 

49. a = 7|, y == 30°. 51. p .= 6, y = 45°. 

50. a = - 3, (x h y x ) = (2, 5). 52. p = 6, y = 225°. 

Note. The best way, in general, to construct a straight line from its 
equation is to find its intercepts ($ 26), and then lay them off on the 
axes. If the line passes through the origin, it has no intercepts ; but 
in this case a second point in the line is easily found by assuming any 
convenient value (as 1) for x, and then computing the corresponding 
value of y from the given equation. 

The intersection of two straight lines is to be found by the general 
method explained in $ 27. 

Construct the following lines, and find the point of inter- 
section : 

53. 5x — 2y + ll=0 and y = *lx. 

54. 3^ + 5y-13 = and 4^-y — 2 = 0. 

55. 2#+3y = 7 and x — y = l. 

56. y + l3 = 5x and y + l9=7x. 

57. 3x + y +12 = and 2y = Sx — 6. 

58. |+|=1 and | + | = 0. 

59. Find the vertices of the triangle whose sides are the 
lines 2^ + 92/ +13 = 0, y = 7x — 38, 2y — x = 2. 



42 ANALYTIC GEOMETRY. 

60. Find the equation of the straight line passing through 
the origin and the intersection of the lines Zx — 2y + 4 = 
and Sx + 4y — 5. Also find the distance between these two 
points. 

61. What is the equation of the line passing through (x x , y^) t 
and equally inclined to the two axes ? 

62. Find the equations of the diagonals of the parallelo- 
gram formed by the lines x = a, x — b, x = c, x = d. 

63. Show that the lines y = 2x + 3, y = 3^ + 4, y = 4^ + 5 
all pass through one point. 

Find the intersection of two of the lines, and then see if its co-ordi- 
nates will satisfy the equation of the remaining line. 

64. The vertices of a triangle are (0, 0), (# 1} 0), (x 2 , y 2 ). 
Find the equations of its medians, and prove that they meet 
in one point. 

65. What must be the value of m if the line y — mx passes 
through the point (1, 4) ? 

66. The line y—mx-{-Z passes through the intersection of 
the lines y = x + 1 and y — 2x + 2. Determine the value of m. 

67. Find the value of b if the line y = 6 x + b passes through 
the point (2, 3). 

68. What condition must be satisfied if the points (.r 1? y^) 1 
(#2, V'i), (^3» 3/3) li- e i n one straight line? 

Hot. Let equation [4] represent the line through (x v y x ) and 
(aj 2 , y 2 ) ; then (x 3 , y 3 ) must satisfy it. 

69. Discuss equation [5] for the following cases : (i.) (x l} y x ) 
= (0,0), (ii.)m = 0, (iii.)m = oo. 

70. Discuss equation [6] for the following cases : (i.) 5 = 0, 
(ii.) m = 0, (iii.) m — 00, (iv.) m = 0, and b — 0. 

71. Discuss equation [7] for the following cases : (i.) a = b, 
(ii.)a = 0, (iii.) a = 00, (iv.)J = oo. 



the straight line. 43 

General Equation of the First Degree. 

42. Every equation involving x and y as variables, which 
can be reduced by algebraic operations to the form 

y = mx + b, 
represents a straight line having for slope the coefficient of x, 
and for intercept on the axis of y the value of the constant term. 

Algebraic operations never change the values of x and y 
which will satisfy an equation ; therefore they cannot change 
the locus represented by the equation. 

The equation y = mx J r b, from the manner in which it was 
established, necessarily holds true for all values of m and b 
from —00 to + oo. Therefore every equation of the form 
y = mx J rb must represent some particular one of the infi- 
nite number of straight lines obtained by giving all pos.- 
sible values to the general symbols m and b ; and the values 
of the two constants in the equation must be the values of 
m and b for the particular line represented by the equation. 

43. Every equation which can be put in the form 

a o 
represents a straight line, having for its intercepts the denomi- 
nators of x and y, respectively. 

The proof is similar to that of the preceding proposition. 

44. Every equation of the first degree, with respect to the 
variables x and y, can be put in the form 

Ax + By + C = 0. [9] 

where A, B, C stand for any numbers, positive or negative, 
entire or fractional, rational or irrational. A and B cannot, 
however, both be ; for if they could, then we should also have 
(7=0, and the equation would vanish entirely. Equation [9] 
is termed the General Equation of the First Degree. 



44 ANALYTIC GEOMETRY. 

45. Every equation of the first degree in x and y represents a 
straight line. 

If we reduce the equation to the form Ax -f By -f- C= 0, 
and solve for y, we obtain 

A C 
y B B 

But the equation has now the form y = mx-\-h\ therefore 
it represents a straight line (§ 42), and the values of the slope 
and the intercept on the axis of y are 

m = - — b = -~ 

B B 

46. To find the intercepts and slope of a straight line, having 
given its equation. 

Method I. Find the intercepts, a and b, as in § 26. Then 

(Fig. 19) m = tan y = - tan (180° -h y) = - tan BAO^ - -. 

Method II. Reduce the given equation to the form [6] ; 
and make m equal to the coefficient of x. Eeduce the equa- 
tion to the form [7], and make a and b respectively equal to 
the denominators of x and y. For these operations amount 
to nothing more than substituting particular values in the 
place of general symbols which from their nature include all 
assignable values whatsoever. 

Note. This mode of obtaining from the equation of a curve the values 
of its constants is sometimes called the Method of Equating Coefficients. 

Method III. Determine the values of m, a, and b, in terms 

of A, B, and C } by the method of equating coefficients. The 

results are A C O 

m = 1 a = 5 b — — -• 

BAB 

Nothing now remains but to reduce the given equation to the 
general form, Ax +By + C= 0, and then substitute the par- 
ticular values of A, B, and C. 



THE STRAIGHT LINE. 45 

47. Find the intercepts and slope of the straight line repre- 
sented by the equation 3x — 3 = 4y + 9. 

By I. If y = 0, x =* 4 ; if x = 0, y = — 3. 

i* 6—33 
Therefore a — V, b = — 3, m = == — = -• 

" a 4 4 

By II. The equation put into the forms [6] and [7] becomes 

y = -x— 3, and — (--£-= 1. 

y 4 4-3 

3 
Therefore a = 4, b = — 3, m = -' 

4 

By III. The given equation in the form [9] is 

3;r-4y — 12 = 0. 

Therefore ^4 = 3, 5 -=-4, (7= -12, and 



a = 


_=i2_4 j_ 


-12 o 

— - q 772, — 


3 _3 


3 4 ' * 


— 4 


-4 A 






Ex. 12. 





Describe the position of the following lines by determining 
the values of a, b, and m. 



1. 


4 T 7 


10. 


x v — \ 

2 3 




2. 


y 3 


11. 


* + y. =_i. 

2 3 




3. 


3z + 2 = 2y 


12. 


3y = a;. 




4. 


4y = 5a:. 


13. 


3x = y. 




5. 


7# + 3y = 0. 


14. 


5ar-4y + 20 = 


= 0. 


6. 


4y = 3a; + 24. 


15. 


y = 6*4-12. 




7. 


# + y = 3. 


16. 


y 4- 2 = a; - 4. 




8. 


4y + rc + ll=0. 


17. 


ar + V3y4-10i 


= 0. 


9. 


5a;- 3y + 15 = 0. 


18. 


a;— V3y-10 = 


= 0. 



46 ANALYTIC GEOMETRY. 

19. Discuss equation [9] for the following cases : 

(i.) A = 0. (iv.) .4 = 00. (vii.) A=B, 0=0. 

(ii.) _£ = 0. (v.) A = C=0. (viii.) A = -B, C=0. 
(iii.) (7-0. (vi.) A = B. 

20. Reduce equation [7] to the form of equation [6], and 
find the value of m in terms of a and b. 

21. What value must C have in order that the line 
4:X — by+C may pass through the origin? Through the 
point (2, 0) ? 

22. Determine the values of A, B, and C, so that the line 
Ax -\- By -\- C = may pass through the points (3,0) and 
(0,-12). 

Since the co-ordinates of the given points must satisfy the equation, 
we have the two relations 3 A + (7= and — 12 B + C= 0. 

23. From equation [9] deduce equation [4] by the method 
used in solving No. 22. 

24. If equations [4] and [9] represent the same line, what 
are the values A, B, C, in terms of x lt y 1} x 2 and y 2 ? 

25. In equation [4] find the values of m and b in terms of 

x ii yu x ii y<L- 

26. In equation [8] find the values of cos a, sin a, and p, in 
terms of the general constants A, B, C. 

To reduce the form [9] to the form [8] we must multiply each term 
by a certain quantity .k ; to determine k, we have 

k(Ax — By + C) = x cos a -f y sin a — p = 0, 
whence, by equating coefficients, we obtain 

cos a = kA, sin a = kB, p = — hC. 
But cos 2 a + sin 2 a = 1, 

whence we have k= ± — . 

VA 2 + B 2 
By substituting the value of k, we have 

A ■ B C 

cos a = , smo= , p = , 

± V^i 2 + £ 2 ± V^l 2 + .g 2 ± V.4 2 + B 2 



THE STRAIGHT LINE. 



47 



and we should choose that sign before the radical which will make p 
positive (§ 41), therefore the sign which is unlike that of C. 

27. What are the values of cos a, sin a, and p for the fol- 
lowing lines ? Construct the lines. Show that they enclose a 
parallelogram. 

(a) 12^ + 5y- 26-0. (c) 12^-5^-26 = 0. 

(Y) 12x + 5t/ + 26 = 0. (d) 12x-5y + 26 = 0. 



Parallels and Perpendiculars. 

48. If the lines represented by the equations y = mx + b 
and y = m'x + b f are parallel, then we must have, by Geom- 
etry, 



and therefore 



y' = y; 



m 



m. 




Fig. 21. 

If the two lines are perpendicular, then, by Geometry, 

y = y + 90°- (see Fig. 21). 

Therefore, by Trigonometry, 

tan y' = — cot y, 

1 

or m f = 1 

m 

or mm' = -l, 



48 ANALYTIC GEOMETRY. 

49. Conversely, prove that the lines represented by the equa- 
tions y — mx -f- b and y — m!x + b f are parallel if wi = m! ; 
perpendicular, if mm) ' = — 1. 

Hint. Use the above reasoning, taken in reverse order. 

Noth The equations m = m' and mm' = — 1 are examples of Equa- 
tions of Condition ; they express the conditions which must be satisfied 
in order that two lines may be parallel or perpendicular, respectively. 

50. To find the equations of a straight line passing through 
the point (x u yf) and (i.) parallel, (ii.) perpendicular, to the 
line y = mx -f- b. 

The slope of the required line is m in case (i.), and 

in case (ii.) ; and in both cases the line passes through a given 
point (x l9 yO. 

Therefore (§ 38) the required equations are 

(1) y-y 1 = m(x — xf), 

(2) y-y 1 = --(x-x 1 ). 

51. We shall now solve the problem of § 50 in another way. 
Let the equation of a given line be 

Ax+£y + C=-0. (1) 

If we change C to any other value K, but leave A and B 
unchanged, the new equation 

Ax + By + K=Q . (2) 

will represent a line parallel to the given line, because the 
two lines have the same slope (§ 49). 

If we change C to K, as before, and also interchange the 
coefficients A and B, and alter the sign of one of them, the new 
equation Bx-Ay + K=Q (3) 

will represent a line perpendicular to the given line, because . 
the slopes of the two lines satisfy the condition mm 1 ~ — 1. 



THE STRAIGHT LINE. 49 

By assigning different values to iT, equations (1) and (2) 
may be made to represent different parallels, and different 
perpendiculars, respectively, to the given line ; and if we 
regard K as entirely undetermined, then equation (2) may 
be said to represent all parallels, and equation (3) all perr 
pendiculars, to the given line. 

But if in either case we assign a particular value to K, or 
add a new condition which determines K, then the equation 
will represent one definite straight line. 

Suppose we add the condition that the line must pass 
through a given point (# 1} y x ) ; then these co-ordinates must 
satisfy equations (2) and (3), and we have 

A^+By^+K^O, and Bx 1 -Aij 1 +K=0. 

Hence K= — (J.# 1 + _Sy 1 ), and K-=Ay x — Bx x . 

Substituting these values of K in (2) and (3), we obtain for 

the equations of the required lines 

Ax + By = Ax l +By 1 . (4) 

Bx — Ay = Bx x — Ay x . (5) 

Ex. 13. 
Find the equation of a straight line 

1. Passing through (3, —7), and II to the line y— 3^ — 5. 

2. Passing through (5, 3), and II to the line ^y — \x = l. 

3. Passing through (0, 0), and i| to the line y — 4 # = 10. 

4. Passing through (5, 8), and || to the axis of x. 

5. Passing through (5, 8), and || to the axis of y. 

6. Passing through (3,-13), and JL to the line y = 4 x — 7. 

7. Passing through (2, 9), and JL to the line 7y+ 23 # — 5 = 0. 

8. Passing through (0, 0), and JL to the line z + 2y = l. 

9. Perpendicular to the line 5x — 7y + l = 0, and erected 
at the point whose abscissa = 1. 



50 



ANALYTIC GEOMETRY. 



10. Perpendicular to the line y — 3x = 2, and passing 
through the intersection of the lines x — y = l and 2x + 

Angles. 

52. To find the angle formed by the lines y = mx + b, and 
y = m } x-\- b\ 

Let AB and CD (Fig. 22), represent the two lines, respec- 
tively, meeting in the point P. 

Let the angle APC=<f>, and tan <f> = t. By Geometry, 
</> = y — y f . Whence, by Trigonometry, 
. m — m f 



1 + mm' 

This equation determines the value of <£. 



[10] 



Y 




B\ 


Id 




*\\/ 




/9\ 





/c ^\x 


. / 





Y 


1 3/ 




]%/ 




Jr 





/ x 


/ 





Fig. 22. 



Fig. 23. 



53. To find the equation of a straight line passing through 
a given point (x X) y^, and making a given angle cj> with a given 
line y = mx-{-b. 

Let the required equation be 

y - y\ *= m \ x — x \) 

where m) is not yet determined. 



THE STRAIGHT LIN^. 51 

Since the required line may lie either as PQ or PR 
(Fig. 23), we shall have (by § 52) r 

. , m! — m m — m f 

tan </> = or 



1 -f- mm) 1 -j- mm 1 

H, m ± tan d> 
ence m = -*— , 

1 ^m tan </> 
and the required equation is 

m ± tan + . . r -- n 

and (as Fig. 23 shows) there are in general two straight lines 
satisfying the given conditions. 

Ex. 14. 

1. Find the angle formed by the lines ^ + 2y + l = and 
x — 3y-4 = 0. 

The two slopes are —\ and J. If we put m = — J, ra ; = J, we obtain 
4-=— 1, £==135°. If weput m = J, m' = — J, weget < = 1, = 45°. Show 
that both these results are correct. 

Find the tangent of the angle formed by the lines 

2. 3^-4y — 7 and 2#-y = 3. 

3. 2^ + 3y + 4 = and 3^ + 4y + 5=:0. 

4. y — nx = l and 2(y — l) = w#. 

Find the angle formed by the lines 

5. x-\-y = \ and y = # + 4. 

6. y + 3 = 2.2: and y + 3^ = 2. 

7. 2*+3y + 7=0 and 5ar — 2y + 4 = 0. 

8. 6a?.==2y + 3 .and y-3rr=10. 

9. ^ + 3 - and y - V3x + 4 = 0. 



52 ANALYTIC GEOMETRY. 

10. Discuss equation [11] for the cases where <£ = 0° and 
<jf> = 90°. 

Note. The learner should try to solve the next five exercises directly, 
without using equation [11] ; then verify the result by means of [11]. 

Find the equation of a straight line 

11. Passing through the point (3, 5), and making the angle 
45° with the line 2#-3y + 5 = 0. 

12. Passing through the point (—2, 1), and making the 
angle 45° with the line 2y = 6 — 3x. 

13. Passing through that point of the line y = 2x — 1 for 
which x — 2, and making the angle 30° with the same line. 

14. Passing through (1, 3), and making the angle 30° with 
the line x-2y + l = 0. 

15. Prove that the lines represented by the equations 

Ax + By + C= 0, A'x + B'y + C = 
are parallel if AB 1 — A'B ; perpendicular, if A A 1 = — BB } . 

16. Given the equation 3^ + 4?/ +6 = 0; show that the 
general equations representing (i.) all parallels and (ii.) all 
perpendiculars to the given line are 

(i.) 3^ + 4y + JT-=0. 
(ii.) 4^-3y + iT=0. 

17. Deduce the following equations for lines passing 
through (#!, y x ) and (i.) parallel, (ii.) perpendicular, to the 
line y = mx -f- b. 

(i.) y — mx = y 1 ~ mx Y . 
(ii.) my -f x = my x + a^. 

18. "Write the equations of 3 lines parallel, and 3 lines per- 
pendicular, to the line 2^ + 3y + l = 0. 



THE STRAIGHT LINE. 53 

19. Among the following lines select parallel lines ; per- 
pendicular lines ; lines neither parallel nor perpendicular : 

(i.) 2x + Zy-l=0. (v.) x-y = 2. 

(ii.) 3;r-2y = 20. (vi.) 50 + y)- 11=0. 

(iii.) 4# + 6y = 0. (vii.) x = 8. 

(iv.) 12a; = 8y + 7. (viii.) y + 10 = 0. 

20. Prove that the angle <£, between the lines 

Ax + By + C=0 and A'x + £'y + C = 0, 
is determined by the equation 

tan d> = -• 

21. From the preceding equation deduce the conditions of 
parallel lines and perpendicular lines given in No. 15, p. 52. 

Find the equation of a straight line 

22. Parallel to 2x -f- 3y + 6 = 0, and passing through (5, 7). 

23. Parallel to 2x-\-y — 1 = 0, and passing through the 
intersection of 3#-f-2y — 59 = and 5x — 7y + 6 = 0. 

24. Parallel to the line joining (—2, 7) and (—4,-5), 
and passing through (5, 3). 

25. Parallel to y = mx-{-h, and at a distance d from the 
origin. 

26. Perpendicular to Ax-{-By-\-C= 0, and cutting an in- 
tercept b on the axis of y. 

x 1/ 

27. Perpendicular to - +t = 1, and passing through (a, h). 

x v 

28. Making the angle 45° with - ^f j = 1 ; and passing 

through (a, 0). 

29. Show that the triangle w T hose vertices are the points 
(2, 1), (3, -2), (-4, -1) is a right triangle. 



54 ANALYTIC GEOMETRY. 

30. The vertices of a triangle are (—1,-1), (—3,5), 
(7, 11). Find the equations of its altitudes. Prove that 
the altitudes meet in one point. 

31. Find the equation of the perpendicular erected at the 
middle point of the line joining (5, 2) to the intersection of 
x + 2y — 11 = and 9x — 2y + 59 = 0. 

32. Find the equations of the perpendiculars erected at the 
middle points of the sides of the triangle whose vertices are 
(5, —7), (1, 11), ( — 4, 13). Prove that these perpendiculars 
meet in one point. 

33. The equations of the sides of a triangle are 
^+y+l=0, 3^ + 5y + ll=0, ^ + 2y + 4 = 0. 

Find (i.) the equations of the perpendiculars erected at the 
middle points of the sides ; (ii.) the co-ordinates of their com- 
mon point of intersection ; (iii.) the distance of this point from 
the vertices of the triangle. 

34. Show that the straight line passing through (a, b) and 
(<?, d) is perpendicular to the straight line passing through 
(b, —a) and (c?, — c). 

35. What is the equation of a straight line passing through 
( x i, Vi)> an d making an angle <£ with the line Ax-\-By-\-C~Q ? 

Distances. 

54, Find the distance from the point ( — 4, 1) to the line 
3x — 4y + l=0. Ans. 3. 

The required distance is the length of the perpendicular let 
fall from the given point to the given line. The first method 
that occurs for solving the problem is to form the equation of 
this perpendicular, find its intersection with the given line, 
and then compute the distance from this intersection to the 
given point. 

Let this method be followed in solving the above problem 
and the first five problems of Ex. 15. 



THE STRAIGHT LINE. 



55 



55. To find the distance from the point (x X) y x ) to the line 

Ax + By + C=0. 

Let P (Fig. 24) represent the given point (x l} y x ), and AB 
the given line Ax + By + C = 0. Draw PS J- to AB, and 
PM _L to the axis of x and meeting AB in a point P. Let 
c? denote the required distance PS. Then, by Trigonometry, 

d=PR cos i2P# - PR cos &4X= Pi? cos y. 





Y 


3<i 


-<v 




! Q 


/ 





M X 






Fig. 24. 



Since P is in the given line, and OM= x lt 

-Ax x -G 



PM-- 



B 



Therefore PR = MP - MR = A *i+2?/i + C 

B 



To find the value of cos y, we may use the relations 

A ' 

cosy B 



2 , • 2 i j sin 7 ^ 

cos y + srn\y = 1, and - = — 



Eliminating sin y, we obtain 
B 



cosy = 



^A 2 +B 2 

Substituting these values of PR and cosy, we have 
Ax, + By x + C 



d 



VA 2 + B 2 



[12] 



56 ANALYTIC GEOMETRY. 

So long as we are concerned with a single distance, there is 
no occasion for the use of both signs, and we should choose 
that sign which will make d positive. 

Hence, to find the distance from the point (x x , y x ) to the line 
Ax -f By + C = 0, we have as a practical rule : Write x\ for 
x and y x for y, and divide the value of the resulting expression 
by s/A' + B 2 . 

Solve this problem when the given point is assumed to be 
at Q (see Fig. 24) on the same side of the line as the origin. 

What is the value of d when the given point is (i.) the 
origin, (ii.) in the given line ? 

56. To find the distance from the point (z 1} y x ) to the line 
x cos a + y sin a — p = 0. 

Let the equation of the line through the given point, par- 
allel to the given line, be (§ 41) 

x cos a -f- y sin a — p l = 0. 

Since (x x , y x ) is in this line, 

x 1 cos a + y x sin a—p l = 0. 
Therefore p 1 = x x cos a -f y 1 sin a. 

Now p and p f are the distances from the origin to the given 
line, and its parallel, respectively ; therefore, if d denote the 
required distance, 

d = ± (j yl ~~P) ~ — (^i cos a + 3/i sm a ~P) 5 
and in general we should choose that sign which will make d 
positive : the positive sign, if (x 1} t/j) and the origin are on 
opposite sides of the given line ; the negative sign, if they are 
on the same side. 

If, then, the equation of a straight line is reduced to the 
normal form x cos a + y sin a — p = 0, the distance from any 
point (xi, yi) to the line is found simply by substituting on the 
left-hand side of the equation x x for x and y x for y, and then 
computing the value of the expression. 



THE STRAIGHT LINE. 



57 



Ex. 15. 

1. Find the distance from (1, 13) to the line Sx = y — 5. 

2. Find the distance from (8, 4) to the line y = 2x — 16. 

3. Find the distance from the origin to the line 3#+4y=20. 

4. Find the distance from (2, 3) to the line 2# + y — 4 = 0. 

5. Find the distance from (3, 3) to the line y = 4:r— 9. 

6. Prove that the distance from the point (x u y x ) to the line 

y = mx + b is -, 



Vl + ra a 

that sign being chosen which will make d positive. Express 
this result in the form of a rule for practice. 

7. Find the distances from the line 3# + 4y + 15 = to 
the following points : (3, 0), (3, - 1), (3, - 2), (3, - 3), (3, - 4), 
(3, -5), (3,-6), (3, -7), (0, 0), (-1, 0), (-2, 0), (-3, 0), 
(-4,0), (-5,0), (-6,0). 

8. Find the distances from (1, 3) to the following lines : 



337 + 4?/+ 15 = 0. 
3* + 4y+10 = 0. 
3.r + 4y+ 5 = 0. 
3s + 4y =0. 



3:r + 4y- 5 = 0. 
3^ + 4y-10 = 0. 
3# + 4y-15 = 0. 
3^ + 4y-20 = 0. 



Find the following distances : 

9. From the point (2, — 5) to the line y — 3x = 7. 

10. From the point (4, 5) to the line 4y + hx = 20. 

11. From the point (2, 3) to the line x + y = 1. 

12. From the point (0, 1) to the line Sx — 3y = l. 

13. From the point (—1, 3) to the line 3^ + 4y+2 = 0. 



58 ANALYTIC GEOMETRY. 

14. From the origin to the line 3# + 2y — 6 = 0. 

15. From the point (2, —7) to the line joining (—4, 1) and 
(3, 2). 

16. From the line y — 7x to the intersection of the lines 
y = 3 x — 4 and y = 5 x + 2. 

17. From the origin to the line a(x — a) + £(# — S) = 0. 

18. From the points (a, 5) and (— a, — J) to the line 

a 6 

19. From the point (a, b) to the line a# + by = 0. 

20. From the point (A, 4) to the line .4# + By + (7 = Z>. 

Find the distance between the two parallels : 

21. 3s; + 4y + 15 = and 3x + ±y + 5 = 0. 

22. 3# + 4y+15 = and 3* + 4y — 5 = 0. 

23. Ax + JBy + C=0 and Ax + JBy + C' = 0. 

24. 9.r + 3y-7=0 and 9^ + 3y-27=0. 

25. y = bx — 7 and y = 5# + 3. 

x,y n .XV 1 

26. - + T = 2 and -+f=3- 

27. Show that the locus of a point which is equidistant 
from the lines 3^ + 4y — 12 = and 4^ + 3y — 24 = con- 
sists of two straight lines. Find their equations, and draw a 
figure representing the four lines. 

28. Show that the locus of a point which so moves that 
the sum of its distances from two given straight lines is con- 
stant is a straight line. 



the straight line. 59 

Areas. 

57. Find the area of the triangle whose vertices are the 
points (2, 1), (3, -2), (-4, -1). A ns. 10. 

It is shown in Elementary Geometry that the area of a 
triangle is equal to one-half the product of its base and its 
altitude ; hence this problem may be solved by performing 
the following operations : 

(i.) Find the length of one side chosen as base, 
(ii.) Find the equation of the altitude, 
(iii.) Find the intersection of the base and the altitude, 
(iv.) Find the length of the altitude, 
(v.) Multiply this length by one-half the base. 

Let the first five problems of Ex. 16 be solved in this way. 

58. Find the area of a triangle, having given its vertices 
Oi,yi), (> 2 , 2/2), (#3,3/3). 

Solution I. If we take as base the line joining (x Y y Y ) 
and (x % , y 2 ), then 

base = V(y 2 - yj 2 + O2 — ^i) 2 . 

The altitude is the distance from (x 3 , y 3 } to the base. 
Writing in equation [12] (p. 55) x 3 for x lf y 3 for y u and for 
A, B,C the values 

A = y* — Un B = — {x 2 — x 1 ) i C = x 2 y l ~x l y 2 , 
obtained by equating co-efficients in [4] and [9] (pp. 37, 43), 
we get 

altitude = ^ ~ y >» _ (^ - *Qy» + * & - *#. . 

"v(y 2 - yd* + Oi - %if 

The area of the triangle = \ base X altitude ; therefore 
area = \ [(y 2 — y x )x z — (x 2 — x,)y 3 + x 2 ij Y — x<y 2 \ 
which may be written more symmetrically thus : 

area = \ [x 1 (y 2 — y z ) + x 2 (y d — y Y ) + x z {y x — y 2 )\. [13] 



60 



ANALYTIC GEOMETRY. 



Solution II. Let PQP (Fig. 25) be the given triangle, 
and let the co-ordinates of PQP be x Y y x , x 2 y 2l % 3 y z , respectively. 
Drop the perpendiculars P3L, QN, PL ; then 

area PQP = PQNM+ PLNQ - PMLP. 
By Geometry, 

PQNM= \MN(PM+ QN) 

= K*2 - x i) (y 2 + 2/1). 

Similarly. 

PLNQ = %(x 3 — x 2 ) (3/3 + y 2 ), 
PMLP = ^x s -x 1 y(ys + yi^ 

Substituting these values, we have 

area PQP == \ [(x 2 — xj (y 2 + y Y ) + (x z — x 2 ) (y 3 + y 2 ) 

- O3 - x,) (y 3 + yO] 
= h [ x &i — ^13/2 + x$2 — ^23/3 + *&z — ^33/1] 

= i[*i(y 2 — y 3 ) + ^2(3/3 - 2/1) + a7 3 (yi - 3/2)]. 



F 


/ 


% 


B 





M I 


* 1 


,X 



Fig. 25. 



Ex. 16. 

Find the area of the triangle whose vertices are the points : 

1. (0, 0), (1, 2), (2, 1). 

2. (3,4), (-3,-4), (0,4). 

3. (2,3), (4,-5), (-3,-6). 

4. (8,3), (-2,3), (4,-5). 

5. (a,0), (-a.0), (0,6). 



THE STRAIGHT LINE. 61 

6. Compare the formula for the area of a triangle with the 
result obtained by solving No. 68, p. 42. What, then, is the 
geometric meaning of that result ? 

Find the area of the figure having for vertices the points : 

7. (3, 5), (7, 11), (9, 1). 

8. (3,-2), (5, 4), (-7, 3). 

9. (-1,2), (4,4), (6,-3). 

10. (0, 0), (x x , 2/0, (x 2 , yj. 

11. (2,-5), (2,8), (-2,-5). 

12. (10,5), (-2,5), (-5,-3), (7,-3). 

13. (0, 0), (5, 0), (9, 11), (0, 3). 

14. (a, 1), (0, b), (c, 1). 

15. (a. &), (b, a), (c, c). 

16. (a, i), (5, a), (c, —c). 

17. Find the angles and the area of the triangle whose 
vertices are (3, 0), (0, 3V3), (6, 3 V3). 

What is the area contained by the lines 

18. ar — 0, y = 0, 5:r + 4y = 20? 

19. x + y = l, x — y = 0, y = 0? 

20. x + 2y = 5, 2x + y = 7, y = x + l? 

21. #-fy = 0, # = y, y=3a? 

22. y = 3x, y = 7x, y=c? 

23. x = 0, y = 0, a?r-4==A y + 6 = 0? 

24. 3^ + y + 4-0, 3^-5y + 34 = 0, 3x-2y + l = 0? 

25. ^_5y + 13 = 0, 5^ + 7y + l-0, 3ar+y-9 = 0? 

26. a; — y = 0, a; + y = 0, x — y = a } x + y = b? 



62 ANALYTIC GEOMETRY. 

Find the area contained by the lines : 

27. x = 0, y = 0, y — mx-{-b. 

28. x = 0, y = 0, - + ? = 1. 

29. a? = 0, y = 0, .4a;+.By + C=0. 

30. y = 3^ — 9, y = 3^ + 5, 2y = ^— 6, 2y = s; + 14. 

31. What is the area of the triangle formed by drawing 
straight lines from the point (2, 11) to the points in the line 
y = 5 x — 6 for which x x = 4, x 2 = 7 ? 

Ex. 17. (Review.) 

1. Deduce equation [7], p. 39, from equation [6]. 

2. The equation y = mx + & is not so general as the equa- 
tion Ax-\-By-\-C— 0, because it cannot represent a line 
parallel to the axis of y. Explain more fully. 

Determine for the following lines the values of a, b, y, p, and a : 

3. a? =2. 6. ^ + V3y = 2. 

4. x = y. 7. ;r — Voy — 2. 

5. y + l = V3(# + 2). 8. V3.r-y = 2. 

9. Find the equations of the diagonals of the figure formed 
by the lines 3 a; — y + 9 = 0, 8x = y — 1, 5#-f-3y = 18, 
5 x -\-Zy- 2. What kind of quadrilateral is it? Why? 

10. Find the distance between the parallels 9x — y + 1 
and 9x = y — 7. 

11. The vertices of a quadrilateral are (3, 12), (7, 9), 
(2, — 3), (— 2, 0). Find the equations of its sides and its area. 

12. The vertices of a quadrilateral are (6, —4), (4, 4), 
(—4, 2), (—8, —6). Prove that the lines joining the middle 
points of adjacent sides form a parallelogram. Find the area 
of this parallelogram. 



THE STRAIGHT LINE. 63 

Find the equation of a line passing through (3, 4), and also 

13. Perpendicular to the axis of x. 

14. Making the angle 45° with the axis of x. 

15. Parallel to the line 5# + 6y + 8 = 0. 

1G. Intercepting on the axis of y the distance — 10. 

17. Passing through the point half way between (1, —4) 
and (-5, 4). 

18. Perpendicular to the line joining (3, 4) and (—1, 0). 

Find the equations of the following lines : 

19. A line parallel to the line joining (x lt y x ) and (x 2 , y 2 ), 
and passing through (# 3 , y 3 ). 

20. The lines passing through (8, 3), (4, 3), (- 5, — 2). 

21. A line passing through the intersection of the lines 
2# + 5y + 8 = and 3^ — 4^—7=0, and _L to the latter 
line. 

22. A line J_ to the line Ax — y = 0, and passing through 
that point of the given line whose abscissa is 2. 

23. A line || to the line Sx + 4y = 0, and passing through 
the intersection of the lines x — 2y — a = and x-{-Sy — 2a = 0. 

24. A line through (4, 3), such that the given point bisects 
the portion contained between the axes. 

25. A line through (x u y^, such that the given point bisects 
the portion contained between the axes. 

26. A line through (4, 3), and forming with the axes in the 
second quadrant a triangle whose area is 8. 

27. A line through (4, 3), and forming with the axes in the 
fourth quadrant a triangle whose area is 8. 

28. A line through (—4, 3), such that the portion between 
the axes is divided by the given point in the ratio 5 : 3. 



04 ANALYTIC GEOMETRY. 

29. A line dividing the distance between (—3, 7) and 
(5, —4) in the ratio 4:7, and J- to the line joining these 
points. 

30. The two lines through (3, 5) making the angle 45° w r ith 
the line 2x — 3y-7-0. 

31. The two lines through (7, —5) which make the angle 
45° with the line 6x — 2y + 3 = 0. 

32. The line making the angle 45° with the line joining 
(7, — 1) and (—3, 5), and intercepting the distance 5 on the 
axis of x. 

33. The two lines which pass through the origin and tri- 
sect the portion of the line x + y = 1 included between the 
a'xes. 

34. The two lines || to the line 4.r + 5y + ll = 0, at the 
distance 3 from it. 

35. The bisectors of the angles contained between the lines 

y = 2x J r 4:, y = ox-\-6. 

Hint. Every point in the bisector of an angle is equidistant from 
the sides of the angle. 

36. The bisectors of the angles contained between the lines 
2x — 5y = 0, 4:X + Sy = l2. 

37. The two lines which pass through (3, 12), and whose 
distance from (7, 2) is equal to V58. 

38. The two lines which pass through (—2, 5), and are 
each equidistant from (3, —7) and (—4, 1). 

Find the angle contained between the lines : 

39. y + S = 2x and y + 3x = 2. 

40. y = 5x—7 and 5y + x—3 = Q. 



THE STRAIGHT LINE. 65 

Find the distance : 

41. From the intersection of the lines 3# + 2y + 4 = 0, 
2x + 5y + 8 = to the line y = bx + 6. 

42. From the point (h, k) to the line - + | = 1. 

43. From the origin to the line lix -f- ley = c 2 . 

44. From the point (a, 0) to the line y = mx -| 

Find the area included between the following lines : 

45. x = y, # + y = 0, %=c. 

46. x J ry = Jc, 2 .r = y + 1c, 2 y = # + £. 



x 



y 



47 - 7. + i = l > v = 2 * + b > *-=2y + 



a ' b 



a. 



48. y = 4:X + 7 and the lines which join the origin to those 
points of the given line whose ordinates are — 1 and 19. 

49. The lines joining the middle points of the sides of the 
triangle formed by the lines x — 5y-f-ll=--0, ll# + 6y— -1 = 0, 

* + y + 4 = 0. 

50. Find the area of the quadrilateral whose vertices are 
(0, 0), (0, 5), (11, 9), (7, 0). 

51. "What point in the line 5x — 4y — 28 = is equidis- 
tant from the points (1, 5) and (7, — 3) ? 

52. Prove that the diagonals of a square are _L to each 
other. 

53. Prove that the line joining the middle point of two sides 
of a triangle is parallel to the third side. 

54. What is the geometric meaning of the equation 
xy = 0? 



66 ANALYTIC GEOMETRY. 

55. Show that the three points (3 a, 0), (0, 3 6), (a, 25) 
are in a straight line. 

56. Show that the three lines 5^ + 3y — 7=0, 3^~4y 
— 10 = 0, and x -f- 2y = meet in a point. 

57. What must be the value of a in order that the three 
lines 3x + y-~2 = 0, 2x — y — 3 = 0, and ax + 2y — 3 = 
may meet in a point ? 

What straight lines are represented by the equations : 

58. x 2 + (a-b)x-ab = 0? 

59. xy + bx + ay + ah = ? 

60. # 2 y = a;y 2 ? 

61. 14^ 2 -5^y-y 2 = 0? 

In the following exercises prove that the locus of the point 
is a straight line, and obtain its equation. 

62. The locus of the vertex of a triangle having the base 
and the area constant. 

63. The locus of a point equidistant from the points {x x , y x ) 
and (x.,, y 2 ). 

64. The locus of a point at the. distance d from the line 
Ax + By + C = 0. 

65. The locus of a point so moving that the sum of its 
distances from the axes shall be constant and equal to h. 

66. The locus of a point so moving that the sum of its 
distances from the lines Ax+JBy+C=0, A l x+B'y+C , = 
shall be constant and equal to h. 

67. The locus of the vertex of a triangle, having given the 
base and the difference of the squares of the other sides. 



THE STRAIGHT LINE. 67 

SUPPLEMENTARY PROPOSITIONS. 
Lines passing through One Point. 

59. If S = 0, S 1 —- represent the equations of any two loci 
with the terms all transposed to the left-hand side, and h de- 
notes an arbitrary constant, then the locus represented by the 
equation S + kS ' — passes through every point common to 
the two given loci. 

For it is plain that any co-ordinates which, satisfy the equa- 
tion S=0, and also satisfy the equation S ! = 0, must like- 
wise satisfy the equation jS-\-Jc8' = Q. 

For what values of 1c will the equation S + kS f = repre- 
sent the lines jS=0 and aS" = 0, respectively? 

60. Find the equation of the line joining the point (3, 4) to 
the intersection of the lines 

3a;-2y + 17=0 and ^ + 4y-27—0. 

The method of solving this question which first occurs is 
to find the intersection of the given lines and then apply 
equation [4], p. 37. 

Another method, almost equally obvious, is to employ equa- 
tion [5], which gives at once 

y — 4 = m(x — 3), 
and then determine m by substituting for x and y the co-ordi- 
nates of the intersection of the given lines. 

The following method, founded on the principle stated in 
§ 59, is, however, sometimes preferable, on account of its 
generality and because it saves the labor of solving the given 
equations. According to this principle, the required equation 
may be immediately written in the form 

3:*; - 2y + Yl+k(x + 4y - 27) = 0. 



68 ANALYTIC GEOMETRY. 

And since the line passes through (3, 4), we must have 
9 _ 8 + 17 + k(3 + 16 - 27) = 0, 

whence k = -• 

4 

Therefore 12^-8y+68 + 9^ + 36y-243 - 0, 
or 3o; + 4y-25 — 0. 

This is the equation of the required line. 

61. If the equations of three straight lines are 

Ax + By +.0=0, A t x+B t y + C , = 0, A n x+B"y + C" = 0, 
and roe can find three constants, I, on, n, so that the relation 
l(Ax+By+C) + m(A'x+B l y+C , ) + n(A"x+B"y+C") = 
is identically true, that is, true for all values of x and y, then 
the three lines meet in a point. 

For if the co-ordinates of any point satisfy any two of the 
equations, then the above relation shows that they will also 
satisfy the third equation. 

62. To find the equation of a straight line passing through 
the intersection of the two lines 

Ax + By + C=0, A'x + B'y + C' = 0, 
and bisecting the angle between them. 

There are evidently two bisectors : one bisecting the angle 
in which the origin lies ; the other bisecting the supplementary 
angle. 

The simplest way to obtain their equations is to express 
algebraically the fact (proved in Geometry) that any point 
(x, y) of the bisector is equidistant from the sides of the angle. 
Hence from equation [12], p. 55, we immediately obtain the 
equation An+By + c _ ^ A'oc+B ' y + O f 

which represents both bisectors if we use both signs on the 
right-hand side. 



THE STRAIGHT LINE. 69 

In order to distinguish between the bisectors, it is necessary 
to pay attention to the sign of the distance from a straight 
line to a point. We see from equation [12] that this dis- 
tance changes sign when the point crosses the line ; let it be 
agreed that distances measured to points on the origin side of 
the line shall be considered positive, and that distances meas- 
ured to points on the side remote from the origin shall be 
considered negative. 

Now the distance from the line Ax-\~By-\-C—Q to the 
origin itself is p 



-VA 2 +£ 2 

and in order that this may be always positive, we must place 
before it the same sign as that of 0. It follows that equation 
[14] will represent the bisector of the angle in which the origin 
lies if we choose that sign which will make the two constant 
terms alike in sign. 

If we choose the other sign, the equation of course will 
represent the bisector of the supplementary angle. 

63. To find the equation of a straight line passing through 
the intersection of the two lines 

X cos a + y sin a — p = 0, X COS a ! + y sin o! — p' = 0, 
and bisecting the angle between them. 

Taking the angle which includes the origin, and denoting 
by (x, y) any point in the bisector, we have immediately for 
its equation 

(x cos a + y sin a — p) — (x cos a' -f- y sin a —p') = 0. 

The equation of the bisector of the supplementary angle is 

(x cos a + y sin a — p) + (# cos a' -f- y sin a — p f ) = 0. 

It may be shown from the form of these equations that the 
two bisectors are perpendicular to each other. 



70 ANALYTIC GEOMETRY. 

Ex. 18. 

Find the equation of a line passing through the intersection 
of the lines 3^ + 2y + 17 = 0, # + 4y-27 = 0, and 

1. Passing also through the origin. 

2, Parallel to the line ^ + 2y + 3 — 0. 

• 3. Perpendicular to the line 6x — by = 0. 

4. Equally inclined to the two axes. 

5. Find the equation of a line parallel to the line x == y, 
and passing through the intersection of the lines y = 2x -f- 1 
and y + Sx = 11. 

6. Find the equation of a straight line joining (2, 3) to the 
intersection of the lines 

2x + 3y + l = and 3^-4y=:5. 

7. Find the equation of a straight line joining (0, 0) to the 
intersection of the lines 

5x — 2y + 3 = and 13a? + y='l. 

8. Find the equation of a straight line joining (1, 11) to the 
intersection of the lines 

2x+by-8 = and 3^-4y=8. 

Find the equation of the straight line passing through the 
intersection of the lines Ax -{- By -{- C = and A'x-\-B'y 
+ C = } and also 

9. Passing through the origin. 

10. Drawn parallel to the axis of x. 

11. Passing through the point (x^ y r ). 

12. Find the equation of a straight line passing through 
the intersection of 5x — 4y + 3 = and 7# -f-Hy — 1 = 0, 
and cutting on the axis of y an intercept equal to 6. 



THE STRAIGHT LINE. 71 

13. Find the equation of a straight line passing through 
the intersection of y = 7x — 4 and y = — 2 # + 5, and forming 
with the axis of x the angle 60°. 

14. The distance of a straight line from the origin is 5 ; 
and it passes through the intersection of the lines Sx — 2y 
4-11 = and 6^; + 7y — • 55 = 0. What is its equation ? 

15. "What is the equation of the straight line passing 
through the intersection of bx + ay = ab and y = mx, and 
perpendicular to the former line ? 

Prove that the following lines are concurrent (or pass 
through one point) : 

16. y = 2x + 1, y^^ + 3, y= — 5^ + 15. 

17. 4^-2y-3 = 0, Sx-y + i = 0, 5x — 2y — 1 = 0. 

18. 2x — y = 5, 3x — y = 6, 4x — y = 7. 

19. What is the value of m if the lines 

x y _ x r y _ 

- + y = l, --f-=:l y = mx 

a b b a * 

meet in one point ? 

20. When do the straight lines y = mx -{- b, y ~ mix + & f , 
y — 7?2 r '^ + b" pass through one point ? 

21. Prove that the three altitudes of a triangle meet in one 
point. 

22. Prove that the perpendiculars erected at the middle 
points of the sides of a triangle meet in one point. 

23. Prove that the three medians of a triangle meet in one 
point. Show also that this point is one of three points of 
trisection for each median. 

24. Prove that the bisectors of the three angles of a triangle 
meet in one point. 



72 ANALYTIC GEOMETRY. 

25. The vertices of a triangle are (2, 1), (3,-2), (- 4, - 1). 
Find the lengths of its altitudes. Is the origin within or 
without the triangle ? 

26. The equations of the sides of a triangle are 

3>x + y + 4: = 0, 3a; — 5y + 34 = 0, Sx — 2y + l=0. 
Find the lengths of its altitudes. 

What are the equations of the lines bisecting the angles 
between the lines 

27. Sx — 4y + 7 = and ±x— 3y + 17=0? 

28. 3^-4y-9 = and 12s + 5y-3 = 0? 

29. y = 2^ — 4 and 2y = a; + 10? 

30. x + y = 2 and a; - y = ? 

31. y = m# + 5 and y = ra f # + 5' ? 

32. H-ove that the bisectors of the two supplementary angles 
formed by two intersecting lines are perpendicular to each 
other. 

Equations representing Straight Lines. 

64. A homogeneous equation of the nth degree represents n 
straight lines through the origin. 

Let the equation be 

Ax n + Bx n ~ l y + Cx n ~y + + Ky n — 0. 

Dividing by Ay n , we have 

eH(fw + + f=°- 

If ?'i, r 2 , 1*3, t % h denote the roots of this equation, then the 

equation, resolved into its factors, becomes 



THE STRAIGHT LINE. 73 

and therefore is satisfied when each one of these factors is 
zero, and in no other cases. 

Therefore the locus of the equation consists of the n straight 

x — r 1 y = 1 x — r 2 y = 0, , x — r n y = 0. 

65. To find the angle between the two straight lines repre- 
sented by the equation Ax 2 -j- Hxy -\-By 2 — 0. 

Solving the equation as a quadratic in x, we obtain 

2i^ + (iT±VF-4i% = 0. 
Hence the slopes of the two lines are 

2A , 2A 



-H--y/H 2 -±AB -H+^/H 2 -±AB 

Therefore 

, y/H* — ±AB , A 

m — m' = » mm' = — ; 

B B' 

and (equation [10], p. 50) 

,_ m — m f _ ViP — 4 AB 
\-\-mrn) A-\-B 

66. To find the condition that the general equation of the 
second degree may represent two straight lines. 

We may write the most general form of the equation of the 
second degree as follows : 

Ax 2 + Hxy+By 2 + Dx + Ey + 0= 0. (1) 

In order that this equation may represent two straight lines, 
it must be equivalent to the product of two linear factors ; 
that is, equivalent to an equation of the form 

(Ix + my + n) (px -f qy + r) = 0. (2) 

Equating coefficients in (1) and (2), we obtain 

lp = A, mq=B, nr=C, 

Iq + mp —H, lr-\-np-= I), mr -\-nq = E. 



74 ANALYTIC GEOMETRY. 

The product of IT, D, and E is 

BEE = 2 Imnpqr + lp(rt 2 q 2 + m 2 r 2 ) + mq(l 2 r 2 + n 2 p 2 ) 
+ nr{m l p 2 + f'g' 2 ) 
-2 ^^(7 +J.(^7 2 - 2BC) + B(E 2 - 2 AC) 
+ C(B 2 -2AB). 
Hence the required condition is 

4 ABC- AE 2 - BE 2 - CH 2 + BEE = 0. 

Ex. 19. 

1. Describe the position of the two straight lines repre- 
sented by the equation Ax 2 -f- Bxy + By 2 -f Dx + i?y -f (7= 0, 
where (i.) A = jET= .2) .=== 0, (ii.) B = H=E=0. 

2. When will the equation a^:y + Z># -f cy + cZ — repre- 
sent two straight lines ? 

3. Find the conditions that, the straight lines represented 
by the equation Ax 2 + Bxy -f- By 1 = may be real ; imagi- 
nary ; coincident ; perpendicular to each other. 

4. Show that the two straight lines x 2 — 2xy sec 6 + y 2 = 
make the angle with each other. 

Show that the following equations represent straight lines, 
and find their separate equations : 

5. x 2 -2xy-3y 2 + 2x-2y + l=0. 

6. x 2 — 4 xy + by 2 — 6 y -f 9 — 0. 

7. # 2 -4^ + 3y 2 +6y--9 = 0. 

8. Show that the equation x 2j r xy — 6y 2 +7^ + 31.y— 18 — 
represents two straight lines, and find the angle between them. 

Determine the values of K for which the following equa- 
tions will represent in each case a pair of straight lines. Are 
the lines real or imaginary? 

9. 12^ 2 -10xy + 2y 2 + ll^-5y + jr=0. 



THE STRAIGHT LINE. 75 

10. 12:r + /i% + 2y 2 +ll:r-5y + 2=0. 

11. 12^ 2 + 36^ + iT?/ 2 + 6^+63/ + 3 = 0. 

12. For what value of K does the equation Kxy + bx 
+ 3y + 2 = represent two straight lines ? 

Problems on Loci involving Three Variables. 

67. A trapezoid is formed by drawing a line parallel to the 
base of a given triangle. Find the locus of the intersection of 
its diagonals. 

If ABC be the given triangle, and we choose for axes the 
base AB and the altitude CO, the vertices A, B, C may be 
represented in general by (a, 0), (b, 0), (0, c), respectively. 
The equations of AC said .5(7 are 

X , V _. XV 

- + ^ = 1 and T + £ = l. 

a c be 

Let y = m be the equation of the line parallel to the base, 
and let it cut AC in D, BC in E\ then the co-ordinates of 
D and U, respectively, are 

■ — am \ t [be — bm 
, m 



c J \ c 

Hence the equation of the diagonal BD is 
y cm 

x — b ac — am — be 
and the equation of the diagonal AE is 
y cm 

x—-a be — bm — ac 
If P be the intersection of the diagonals, then the co-ordi- 
nates x and y of the point P must satisfy both (1) and (2) ; 
by solving these equations, therefore, we obtain for any par- 
ticular value of m the co-ordinates of the point P. But what 
we want is the algebraic relation which is satisfied by the 



a) 

(2) 



76 ANALYTIC GEOMETRY. 

co-ordinates of P, whatever the value of m may he. To find 
this, we have only to eliminate m from equations (1) and (2). 
By doing this we obtain 

2cx+(a + h)y=(a+b)c, 



or 



(a + b) c 



+ * = 1. 



We see from the form of this equation that the required 
locus is the line which joins C to the middle point of AB. 

Remark. The above solution should be studied till it is understood. 
In problems on loci it is often necessary to obtain relations which in- 
volve not only the x and y of a point of the locus which we are seeking, 
but also some third variable (as m in the above example). 

In such cases we must obtain two equations which involve x and y 
and this third variable, and then eliminate the third variable ; the 
resulting equation will be the equation of the locus required. 

Ex. 20. 

1. Through a fixed point any straight line is drawn, meet- 
ing two given parallel straight lines in P and Q ; through P 
and Q straight lines are drawn in fixed directions, meeting 
in P. Prove that the locus of P is a straight line, and find 
its equation. 

2. The hypotenuse of a right triangle slides between the 
axes of x and y, its ends always touching the axes. Find the 
locus of the vertex of the right angle. 

3. Given two fixed points, A and P, one on each of the 
axes ; if U and V are two variable points, one on each axis, 
so taken that 0U+ OV=OA+OB, find the locus of the 
intersection of AV and P U. 

4. Find the locus of the middle points of the rectangles 
which may be inscribed in a given triangle. 

5. If PP, QQ 1 are any two parallels to the sides of a given 
rectangle, find the locus of the intersection of PQ and P f Q*. 



CHAPTER III. 
THE CIRCLE. 

Equations of the Circle. 

68. The Circle is the locus of a point which moves so that 
its distance from a fixed point is constant. The fixed point 
is the centre, and the constant distance the radius, of the circle. 

Note. The word " circle," as here defined, means the same thing as 
"circumference" in Elementary Geometry. This is the usual meaning 
of "circle " in the higher branches of Mathematics. 

69. To find the equation of a circle, having given its centre 
(a, b) and its radius r. 




Fig. 26. 



Let C (Fig. 26) be the centre, and P any point (a>, y) of 
the circumference. Then it is only necessary to express bv 
an equation the fact that the distance from P to C is constant, 
and equal to r : the required equation evidently is (§ 6) 

(as — ay 2 +(y — by = r\ [15] 



78 ANALYTIC GEOMETRY. 

If we draw CB II to OX, to meet the ordinate of P, then 
we see from the figure that the legs of the rt. A CPB are 
CB = x-a, PB = y-b. 

If the origin be taken at the centre, then a = b = 0, and the 

equation of the circle is 

X 2 + y 2 ==r 2 9 [ 16 j 

This is the simplest form of the equation of a circle, and 
the one most commonly used. 

If the origin be taken on the circumference at the point A, 
and the diameter AB be taken as the axis of x, then the centre 
will be the point (r, 0). Writing r in place of a, and in place 
of b in [15], and reducing, we obtain 

oc 2 + y 2 = 2rx. [17] 

Why is this equation without any constant term ? 

70. To find the conditio?} that the general equation of the second 
degree > Ax 2 + Hxy + By 2 + Bx + Ey + tf= 0, (1) 

shall represent a circle. 

If possible, let it be the circle whose centre is the point 
(a, b) and whose radius is r. The equation of this circle has 
been found to be 

(x-af + (y-by = r\ 
or x 2 + y 2 -2ax-2by + a 2 + b 2 -r 2 = 0. (2) 

Equating corresponding coefficients and constant terms, we 
have A== i^ £z= i } 11=0, 

2> = - 2a, P=-2b, C = a 2 + b 2 - r\ 
Since, if A — B in equation (1), both A and B can be 
reduced to unity by division, the two conditions necessary in 
order that equation (1) may represent a circle are 

A = B and H= ; 
and the general equation of a circle may be written 

x 2 + y 2 + Bx + Ey + C = . 



THE CIRCLE. 79 

The co-ordinates of the centre and the radius have the 
D * E , 



Ex. 21. 

Find the equation of the circle, taking as origin 

1. The point B (Fig. 26). 

2. The point D (Fig. 26). 

3. The point E (Fig. 26). 

Write the equations of the following circles : 

4. Centre (5, —3), radius 10. 

5. Centre (0, —2), radius 11. 

6. Centre (5, 0), radius 5. 

7. Centre (— 5, 0), radius 5. 

8. Centre (2, 3), diameter 10. 



9. Centre (A, k), radius VA 2 + h 2 . 

10. Determine the centre and radius of the circle 

x 2 + y 2 - 10* + 12y + 25 = 0. 

In this case D^-10, #==12, (7= 25 (see § 70). Therefore a = 5, 
b = -6, r=V25 + 36-25 = 6. 

Determine the centres and radii of the following circles : 

11. x 2 + y 2 -2x-±y = 0. 17. 6x 2 — 2y(7 - 3y) = 0. 

12. 3^ 2 +3?/ 2 -5^-7y+l=:0. 18. x 2 + y 2 =9Ic 2 . 

13. 5; 2 + 3/ 2 -8^=:0. 19. (x+y) 2 +(x — y) 2 =8k 2 . 

14. ^ 2 + y 2 + 8^ = 0. 20. x 2 + y 2 = a 2 + b 2 . 

15. ^ 2 + ?/ 2 -8y = a 21. x 2 + y 2 = Jc(x + Jc). 

16. ^ + y 2 + 8y = 0. 22. # 2 + y 2 = /b-|- £y. 



80 ANALYTIC GEOMETRY. 

23. When are the circles x 2 + y 2 + Dx + Uy + C=0 and 
x 2 + y 2 +D'x+E'y + C = concentric? 

24. What is the geometric meaning of the equation (x — a) 2 

25. Find the intercepts of the circles 

(i.) x 2 + y 2 -8x-8y + 7=0, 

(ii.) ^ 2 + y 2 -8^-83/+16 = 0, 

(iii.) x 2 + y 2 - Sx - 8y + 20 = 0. 

Putting y = in each case, we have in case (i.) x 2 — 8 a; + 7=0, 
whence a; = 1 and 7 ; in case (ii.) x 2 — 8 # + 16 = 0, whence x = 4 ; in 
case (iii.) a; 2 — 8 x + 20 = 0, whence x --= ± V— 4. 

Putting cc = in each case, we obtain for y values indentical with the 
above values of x. 

The geometric meaning of these results is as follows : 

Circle (i.) cuts the axis of x in the points (1, 0), (7, 0), and the axis of 
y in the points (0, 1), (0, 7). 

Circle (ii.) touches the axis of x at (4, 0), and the axis of y at (0, 4). 

Circle (iii.) does not meet the axes at all. 

This is the meaning of the imaginary values of x and y in case (iii.). 

If, however, we wish to make the language of Geometry conform 
exactly to that of Algebra, then in this case we should not say that 
the circle does not meet the axes at all, but that it meets them in imagi- 
nary points ; just as we do not say that the equation x 2 — 8 a; + 20 = 
has no roots, but that it has two imaginary roots. 

Find the centres, radii, and intercepts on the axes of the 
following circles : 

26. x 2 + y 2 — 5x — 7y + Q = 0. 

27. ^ 2 + y 2 -12^-4y + 15 = 0. 

28. ^ 2 + y 2 -4^- 8y = 0. 

29. x 2 + y 2 -6x + Ay + 4: = 0. 

30. o; 2 +2/ 2 + 22^-18y+57=:0. 






THE CIRCLE. 81 

31. Under what conditions will the circle x 2 + y 2 +Dx 
-\-JEy + C=0 (i.) touch the axis of x? (ii.) touch the axis 
of y ? (iii.) not meet the axes at all ? 

32. Show that the circle x 2 + y 2 + 10ar — 10y + 25 = 
touches the axes and lies entirely in the second quadrant. 
Write the equation so that it shall represent the same circle 
touching the axes and lying in the third quadrant. 

33. In what points does the straight line 3# + ?/ = 25 cut 
the circle x 2 + y 2 = 65 ? 

34. Find the points common to the loci x 2 + y 2 — 25 and 
y = 2x~ 4. 

35. The equation of a chord of the circle x 2 + y 2 = 4 is 
y = 2x + 11. Find its length. 

x v 

36. The equation of a chord is - + y- =1: that of the circle 

u a b 

is x 2 + y 2 = r 2 . Find the length of the chord. 

37. Find the equation of a line passing through the centre 
of x 2 -\-y 2 — 6x — 8y = — 21 and perpendicular to x-\-2y = 4. 

38. Find the equation of that chord of the circle x 2 -\-y 2 =l30 
which passes through the point for which the abscissa is 9 
and the ordinate negative, and which is II to the straight line 
4^~5y — 7=0. 

39. What is the equation of the chord of the circle 
x* -j- y 2 — 277 which passes through (3,-5) and is bisected 
at this point ? 

40. Find the locus of the centre of a circle passing through 
the points (#1,3/1) and (x 2 ,y 2 ). 

41. What is the locus of the centres of all the circles which 
pass through the points (5, 3) and (—7, — 6) ? 



82 ANALYTIC GEOMETRY. 

Find the equation of a circle : 

42. Passing through the points (4, 0), (0, 4), (6, 4). 

43. Passing through the points (0, 0), (8, 0), (0, -6). 

44. Passing through the points (-6, -1), (0, 0), (0, —1). 

45. Passing through the points (0, 0), (—8a, 0), (0, 6a). 

46. Passing through the points (2, - 3), (3, - 4), (— 2, — 1). 

47. Passing through the points (1, 2), (1, 3), (2, 5). 

48. Passing through (10, 4) and (17, — 3), and radius = 13. 

49. Passing through (3, 6), and touching the axes. 

50. Touching each axis at the distance 4 from the origin. 

51. Touching each axis at the distance a from the origin. 

52. Passing through the origin, and cutting the lengths 
a, b from the axes. 

53. Passing through (5, 6), and having its centre at the 
intersection of the lines y = Ix — 3, 4y — 3 x = 13. 

. 54. Passing through (10, 9) and (5, 2 — 3V6), and having 
its centre in the line 3x — 2y — 17=0.. 

55. Passing through the origin, and cutting equal lengths 
a from the lines x = y, x -f- y = 0. 

56. Circumscribing the triangle whose sides are the lines 

X 1] 

y = 0, y = mx + b, - + | = 1. 

57. Having for diameter the line joining (0, 0) and (x lt y x ). 

58. Having for diameter the line joining (x lt y^) and (x 2 , y 2 ). 

59. Having for diameter the line joining the points where 
y = mix meets x 2 + y 2 = 2rx. 

60. Having for diameter the common chord of the circles 
x 2 -J- y 1 = r 2 and (x — a) 2 + y 2 = r 2 . 



THE CIRCLE. 



83 



Tangents and Normals. 

71. Let QPQ' (Fig. 27) represent any curve. If the secant 
QPE be turned about the point P until the point Q approaches 
indefinitely near to P, then the ultimate position, TT { \ of the 
secant is called the Tangent to the curve at P. 





Y 






T' 








Q^~ 






R 




n/ 




V\T 







M 


\ XX 


/ 






V 




Fig. 27. 



Fig. 28. 



The tangent TT f is said to touch the curve at P, and the 
point P is called the Point of Contact. 

The straight line PjY drawn from P, perpendicular to 
the tangent TT\ is called the Normal to the curve at P. 

Let the curve be referred to the axes OX, OY, and let M 
be the foot of the ordinate of the point P. Let also the tan- 
gent and the normal at P meet the axis of x in the points 
T, JV, respectively. Then MT is called the Subtangent for the 
point P, and MJSf is called the Subnormal. 

72. To find the equation of a tangent to the circle x 2 -\-y 2 --r 2 i 
at the point of contact (x Xl 3/1). 

Let P (Fig. 28) be the point (x 1% y,), and Q any other point 
(x 2 , 2/2) of the circle. Then the equation of the line PQ is 



*h vL± U/*2 JL>\ 



(1) 



84 ANALYTIC GEOMETRY. 

If now we make this secant become a tangent, by turning it 
about P till P coincides with P, then x 2 '— x\, y 2 = y lt and the 

fraction — — — assumes the indeterminate form -• 



llcLULlUll cXCSO 

X 2 Xi 

But we have not 
Q lie in the circle. 
These conditions 


yet 
are 


introduced the condit 



ions that P a 

= 0. 

have 


nd 


Subtracting, (x 2 — x 2 ) -f- (y 2 2 — y x 2 ) =± 
Factoring, (# 2 — ^) (# 2 + x Y ) + (y 2 - 
Whence, by transposition and division, we 

3/2 V\ x 2 r x \ 


= 0. 



%2 — *\ y<i + 3/1 

And by substitution in (1) the equation of the secant 
becomes V ~ Vi _ ^ + x x 

x — x\~ y 2 + y x 

Now let Q coincide with P, or x 2 ~ x ly y 2 = y Y ; the secant 
becomes a tangent at P, and the equation becomes 

y — yi = _ ^1, 
x — x x y x 

or x Y x + y Y y = x? + y x \ 

And, since ^ + y 2 — r 2 , we obtain 

a*» + 2/i2/ = ** 2 , [18] 

an equation easily remembered from its symmetry and because 
it may be formed from x 2 -f y 2 = r 2 by merely changing x 2 to 
x x x and y 2 to y x y. 

Note. The above method of obtaining the equation of the tangent 
to a circle is applicable to any curve whatever. It is sometimes called 
the secant method. 



THE CIRCLE. 85 

73. To find the equation of the normal through (x Xl y{). 
The slope of the tangent is -• 

Therefore that of the normal will be — (§ 48). 
Hence the equation of the normal is (§ 50) 

x x 
which reduces to the form 

y x oc — oo x y = O. [19] 

Therefore the normal passes through the centre. 
We may also obtain the same equation by proceeding as 
in § 51. 

74. To find the equations of the tangent and normal to the 
circle {x — a) 2 + (y — b) 2 — ** 2 at the point of contact (x 1} y Y ). 

We proceed as in § 72, only now the equations of condition 
which place (x Y , y x ) and (x 2 , y 2 ) on the circle are 

(x v — a) 2 + (y x — b) 2 = r\ 
(x 2 - a y + (y 2 -~by = r\ 

After subtracting and factoring, we have 

(x 2 — x Y ) (x 2 + x x — 2a) + (y 2 — y Y ) (y 2 + y 1 —2b) = 0, 

whence V^ ~ V\ x * ~f~ x \ ~~ " a 

x 2 — x\~ y 2 + y l — 2b 

Hence the equation of a secant through ($ lf y x ) and (x 2l y 2 ) is 

y — Vi == __ x 2 + x l — 2a 
x~x x y 2 + y 1 — 2b 

Making x 2 = x lt and y 2 — y u and reducing, we obtain 

(»! - a) (x - a) + (If! -b)(y-b)= r\ [20] 



86 ANALYTIC GEOMETRY. 

Equation [20] may be immediately formed from [18] by 
affixing — a to the x factors and — b to the y factors, on the 
left-hand side. 

By proceeding as in § 73, we obtain for the equation of 
the normal 

(2/i —b)(x — a\j — (a^ — a)(y — y Y ) =0. [21] 



75. To find the condition that the straight line y = mx + c 
shall touch the circle x 2 + y 2 = r 2 . 

I. If the line touch the circle, it is evident that the perpen- 
dicular from the origin to the line must be equal the radius r 

of the circle. The length of this perpendicular is — =z: 

Vl+m 2 
(§ 55). Therefore the required condition is expressed by the 

equation 

<r = r"(l + ra ). 

II. By eliminating y from the equations 

y = mx -f- c, x 2 -f y 2 = r 2 , 
we obtain the quadratic in x, 

(1 -f- ra 2 )x' 2 -f 2 ma = r 2 — c 2 , 
the two roots of which are 



rac Vr'Yl + ?n 2 ) — c 2 
x — ± ^ — z 

Y-\-m 2 ' 1 + wi 1 

If these roots are real, the line will cut the circle ; if they 
are equal, it will touch the circle ; if they are imaginary, it 
will not meet the circle at all. 



The roots will be equal if Vr 2 (l+ ra 2 ) — c 2 = ; that is, if 
c 2 — r 2 (l + m 2 ), a result agreeing with that previously obtained. 

If in the equation y — rax + c we substitute for c the value 

rVl + m2 > we obtain the equation to the tangent of a circle in 

the useful form , /— ^ r 9 on 

y = mas ± rvl -f- m 2 . |_^J 



THE CIRCLE. 87 

This equation, if we regard m as an arbitrary constant, 
represents all possible tangents to the circle x 2 + y 2 = r 2 . 

Note 1. Method II. is applicable to any curve, and agrees with the 
definition of a tangent given in \ 71. 

Note 2. In problems on tangents the learner should consider whether 
the co-ordinates of the point of contact are involved. If they are, he 
should use equation [18] ; if they are not, then in general it is better to 
use equation [20]. 

Ex. 22. 

1. Explain the meaning of the double sign in equation [22]. 

2. Deduce the equations of the tangent and normal to the 
circle x 2 + y 2 — r 2 , assuming that the normal passes through 
the centre. 

3. Find the equations of the tangent and the normal pass- 
ing through the point (4, 6) of the circle x 2 -\-y 2 = 52. Also 
the lengths of tangent, normal, subtangent, subnormal, and 
the portion of the tangent contained between the axes. 

4. A straight line touches the circle x 2 -f y 2 = r 2 in the 
point (#!, y x ). Find the lengths of the subtangent, the sub- 
normal, and the portion of the line contained between the 
axes. 

5. What is the equation of a tangent to the circle 
x 2 + y 2 = 250 at the point whose abscissa is 9 and ordinate 
negative ? 

6. Find the equations of tangents to x 2 -f- y 2 =10 at the 
points whose common abscissa =1. 

7. Tangents are drawn through the points of the circle 
% ?,J ry 2 = z 25 which have abscissas numerically equal to 3. Prove 
that these tangents enclose a rhombus, and find its area. 

8. The subtangent for a certain point of a circle is 5|- ; the 
subnormal is 3. What is the equation of the circle ? 



88 ANALYTIC GEOMETRY. 

Find the equation of a straight line 

9. Touching rr 2 -j-y 2 = 232 at the point whose abscissa =14. 

10. Touching O - 2) 2 + (y — 3) 2 — 10 at the point (5, 4). 

11. Touching x 2 -\- y 2 — ?>x — 4y = at the origin. 

12. Touching x 2 + y 2 — 14 x — 4 y — 5 = at the point whose 
abscissa is equal to 10. 

What is the equation of a straight line touching the circle 
^ 2 + y 1 = r *> an( i also 

13. Passing through the point of contact (r, 0) ? 

14. Parallel to the line Ax +£y + C=0? 

15. Perpendicular to the line Ax + By -f- C= ? 

16. Making the angle 45° with the axis of re? 

17. Passing through the exterior point (A, 0) ? 

18. Cutting off a triangle of area lc 2 from the axes ? 

19. Find the equations of the tangents drawn from the 
point (10, 5) to the circle x 2 + y 2 ==100. 

20. Find the equations of tangents to the circle x 2 + y 2 
+ 10# — 6y — 2 = and II to the line y = 2x — 7. 

21. Find the lengths of subtangent and subnormal in the 
circle x 2 + y 2 — 14 x — 4 y — 5 for the point (10, 9). 

22. What is the equation of the circle (centre at origin) 
which is touched by the straight line x cos a -f- y sin a —p ? 
What are the co-ordinates of the point of contact ? 

23. When will the line Ax + By + C=0 touch the circle 
x 2 + y 2 = r 2 ? the circle (# — a) 2 + (y — S) 2 = r 2 ? 

24. Find the equation of a straight line touching x 2 -\-y 2 
= ax -\-hy and passing through the origin. 



THE CIRCLE. 89 

Prove that the following circles and straight line touch,- 
and find the points of contact in each case : 

25. x 2 + y 2 + ax + by = and ax + by + a 2 + b 2 = 0. 

26. x 2 + y 2 — 2ax — 2by + b 2 = and x = 2a. 

27. x 2 + y 2 = ax + Jy and ax — by-\-b 2 = 0. 

28. What is the equation of the circle (centre at origin) 
which touches the line y = 2>x — 5 ? 

29. "What must be the value of m in order that the line 
y — mx + 10 may touch the circle x 2 -f- y 2 = 100 ? Show that 
we get the same answer for the line y = mx — 10, and explain 
the reason. 

30. Determine the value of cm order that the line 3x—Ay 
+ e = may touch the circle x^ + y 2 — Sx + 12y — 44— 0. 
Explain the double answer. 

31. What is the equation of the circle having for centre 
the point (5, 3) and touching the line 3^ + 2y — 10 = 0? 

32. What is the equation of a circle whose radius = 10, 
and which touches the line 4# + 3y — 70 = in the point 
(10, 10) ? 

33. About the point (5, 9) a circle touching the line 
4#-f 3y-f- 3 = in the point (—3,3) is described. What 
is its equation ? 

x 1/ 

34. Under what condition will the line --J-^ = 1 touch 

the circle x 2 + y 2 = r* ? 

35. What is the equation of the circle inscribed in the 
triangle whose sides are 

* = 0, y = 0, ?+?=!? 



90 ANALYTIC GEOMETRY. 

36. Two circles touch each other when the distance between 
their centres is equal to the sum or the difference of their 
radii. Prove that the circles 

% 2 + V 1 = ( r + a ) 2 > ( x ~ a 7 + y 2 = r 2 

touch each other, and find the equation of the common tangent. 

37. Two circles touch each other when the length of their 
common chord — 0. Find the length of the common chord of 

and hence prove that the two circles touch each other when 
(a-by = 2r\ 

Ex. 23. (Review.) 

Find the radii and centres of the following circles : 

1. Sx 2 -6x + 3y 2 + 9y-l2 = 0. 

2. 7x 2 + Sy 2 -4:7/-(l-2x) 2 = 0. 

3. y(y~-5) = x(3-x). 



4. Vl + a\x 2 + y 2 ) = 2 b(x + ay). 

Find the equation of a circle : 
5. . Centre (0, 0), radius = 9. 

6. Centre (7, 0), radius = 3. . 

7. Centre (—2, 5), radius = 10. 

8. Centre (3a, 4a), radius = 5a. 

9. Centre (b + c, b — c), radius = c. 

10. Passing through (a, 0), (0, 6), (2a, 2b). 

11. Passing through (0, 0), (0, 12), (5, 0). 

12. Passing through (10, 9), (4, - 5), (0, 5). 

13. Touching each axis at the distance —7 from the origin. 



THE CIRCLE. 91 

14. Touching both axes, and radius = r. 

15. Centre (a, a), and cutting chord = b from each axis. 

16. Passing through (0, 0), and touching y = 2x -f- 3. 

17. Passing through (1, — 3), and touching 2x — y — 4 = 0. 

18. With its centre in the line 5x — 7 y — 8 = 0, and 
touching the lines 2x — y = 0, x — 2y — 6 = 0. 

19. Passing through the origin and the points common to 
the circles x 2 + y 2 — 6x — lOy — 15 = 0, 

x 2 + y 2 + 2x + 4y + 20 = 0. 

20. Having its centre in the line 5x — Sy — 7=0, and 
passing through the points common to the same circles as in 
No. 19. 

21. Touching the axis of x, and passing through the points 
common to the circles 

x 2 + y 2 + 4:x~ Uy - 68 = 0, 
x 2 + y 2 - 6 x - 22y + 30 = 0. 

22. Find the centre and the radius of the circle which 
passes through (9, 6), (10, 5), (3, -2). 

23. What is the distance from the centre of the circle 
passing through (2, 0), (8, 0), (5, 9) to the straight line joining 
(0,-11) and (-16,1)? 

24. What is the distance from the centre of the circle 
^ 2 + y 2 -4^ + 8y = to the line Ax -3?/ + 30 = 0? 

25. What portion of the line y = bx -j- 2 is contained within 
the circle x 2 + y 2 - IZx - 4y — 9 = ? 

26. Through that point of the circle x 2 -\-y 2 =-2b for which 
the abscissa = 4 and the ordinate is negative, a straight line 
parallel to y — Sx — 5 is drawn. Find the length of the inter- 
cepted chord. 



92 ANALYTIC GEOMETRY. 

27. Through the point (x u y{), within the circle x 2 -{-y 2 = r 2 , 
a chord is drawn so as to be bisected at this point. What is 
its equation ? 

28. What relation must exist among the coefficients of the 
equation A(x 2 + y 2 ) + Bx + Ey + C '= 0. 

(i.) in order that the circle may touch the axis of x ? 
(ii.) in order that the circle may touch the axis of y ? 
(iii.) in order that the circle may touch both axes? 

29. Under what condition will the straight line y = mx + C 
touch the circle x 2 + y 2 = 2r# ? 

30. What must be the value of k in order that the line 
3 a; + 4y = k may touch the circle y 2 =10x — x 2 ? 

31. Find the equation of the circle which passes through 
the origin and cuts equal lengths a from the lines x = y, 
x + y = 0. 

32. Find the equations of the four circles whose common 
radius = V2a, and which cut chords from each axis equal 
to 2 a. 

33. Find the equation of the circle whose diameter is the 
common chord of the circles x 2 + y 2 = r 2 , (x — a) 2 -j- y 2 = r 2 . 

Find the equation of the straight line 

34. Passing through (0, 0) and the centre of the circle 

x 2 + y 2 = a(x + y). 

35. Passing through the centres of the circles 

x 2 + y 2 = 25 and a; 2 + y 2 + 6a; - 8y — 0. 

36. Passing through (0, 0) and touching the circle 

x 2 + y 2 - 6 x - 12y + 41 = 0. 

37. Parallel to a;+V%-12) = and touching x 2 +y 2 =100. 






THE CIRCLE. 93 

38. Passing through the points common to the circles 

x 2 + y 2 - 2x— 4y — 20-0, 
x 2 + y 2 - 14a - 16y + 100 = 0. 

39. Prove that the common chord of the circles in No. 38 
is perpendicular to the straight line joining their centres. 

40. Find the area of the triangle formed by radii of the 
circle x 2 -f- y 2 = 169 drawn to the points whose abscissas are 
— 12 and +7 and ordinates positive, and the chord passing 
through the same two points. 

41. Prove that an angle inscribed in a semicircle is a right 
angle. 

42. Prove that the radius of a circle drawn perpendicular 
to a chord bisects the chord. 

43. Find the inclination to the axis of x of the line joining 
the centres of the circles x 2 -\- 2x + y 1 = 0, x 2 -f 2y + y 2 = 0. 

44. Determine the point from which tangents drawn to the 
cycles x 2 + y 2 - 2x — 6y + 6 — 0, 

x 2 + y 2 - 22 y - 20 .r + 52 - 0, 
will each be equal to 4V6. 

45. Find the equations of the circles which touch the 
straight lines 6^ + 7y + 9 = and 7a? + 6y + 3 = 0, and 
the latter line in the point (3, —4). 

Obtain and discuss the equations of the following loci : 

46. Locus of the centres of a circle having the radius r and 
passing through the point (x lf y x ). 

47. Locus of the centre of a circle having the radius r and 
touching the circle (x — a) 2 + {y — b) 2 ~ r 2 . 

48. Locus of all points from which tangents drawn to the 
circle (x — a) 2 + (y — b) 2 = r 2 have a given length t 



94 ANALYTIC GEOMETRY. 

49. Locus of the middle point of a chord drawn through a 
fixed point A of a given circle. 

50. Locus of the point M which divides the chord AC, 
drawn through the fixed point A of a given circle, in a given 
ratio AM: MC = m:n. 

51. Locus of a point whose distances from two fixed points, 
A, B, are in a constant ratio m : n. 

52. Locus of a point, the sum of the squares of whose dis- 
tances from two fixed points, A and B, is constant, and equal 
to h\ 

53. Locus of a point, the difference of the squares of whose 
distances from two fixed points, A, B, is constant and equal 
to h\ 

54. Locus of the middle point of a line of constant length 
d which moves so that its ends always touch two fixed per- 
pendicular lines. 

55. Locus of the vertex of a triangle whose base is fixed 
and of constant length, and the angle at the vertex is also 
constant. 

56. One side, AB, of a triangle is constant in length and 
fixed in position ; another side, AC } is constant in length but 
revolves about the point A. Find the locus of the middle 
point of the third side, BC 

57. Find the locus of the intersections of tangents at the 
extremities of a chord whose length is constant. 

58. Given the equation of a circle x 1 + y 1 = r 2 . If its radii 
are produced, each by a length equal to the abscissa of the 
point where it meets the circle, find the locus of the extremi- 
ties of the radii produced. 



THE CIRCLE. 



95 



SUPPLEMENTARY PROPOSITIONS. 

76. To find the locus of the middle points of a system of 
parallel chords in the circle x 2 -\- y 2 = r 2 . 




Fig. 29. 



Let the equation of any one of the chords (Fig. 29) be 
y=.mx-\- c, and let it meet the circle in the points (x 1} y^ and 

(ar * y,) - x+x 

Then (§§ 37 and 72) m = - ^f^ 2 - (1) 



z %r\-x 2 , 



(2) 



?/i + 2/2 

Let (x, y) be the middle point of the chord ; then 2x- 

2y = yi-\-y 2 (§ 8), and by substitution we have 

x 
m — - 1 

y 

a relation which evidently holds true for the middle points of 
all the chords. Therefore (2) is the equation of the locus. 
If we write (2) in the form 

(3) 



y- 



m 



we see that the locus is a straight line passing through the 
centre, and perpendicular to the chords (§ 48). 

The locus of the middle points of a system of parallel 
chords is called a Diameter of the circle; and the chords 
which it bisects are called the Ordinates of the diameter. 



96 



ANALYTIC GEOMETRY. 



77. Two tangents can be drawn to a circle from any point ; 
and these tangents will be real, coincident, or imaginary, ac- 
cording as the point is outside, on, or inside the circle, respec- 
tively. 




Fig. 30. 

Let the equation of the circle be 

% 2 + y 1 — ?,2 « 

Let (x lt y x ) be the point of contact of a tangent, (A, k) any- 
other point in the tangent. Then (A, k) must satisfy the 
equation of the tangent ; therefore 

xji + yjc = r 2 . 

Also, since (x 1} y x ) is on the circle, 



x? + Vi = r\ 



(1) 
(2) 



Eliminating y l5 we have 



hxi 



(3) 



Since equation (3) is a quadratic equation, there are two 
points the tangents at which pass through (h, k). Solving 
(3), we obtain _ ^ ±try/V+V-f> 

Xl h* + P 

and we see that the values of x x are real, coincident, or imagi- 
nary, according as h 2 + h 2 is greater than, equal to, or less 
than r 2 ; that is to say, according as (h, h) is outside, on, or 
inside the circle. 



THE CIRCLE. 97 

78. Tangents are drawn to the circle x 2 + y 2 — r 2 from any 
point (A, Jc) ; to find the equation of the straight line joining 
the two points of contact. 

Let (#!, yi), (x 2 , y 2 ) be the points of contact ; then the equa- 
tions of the tangents are (§ 72) 

%& + ViV = r\ 
x 2 x + y 2 y == r 2 . 

Since both tangents pass through (A, Jc), both these equa- 
tions are satisfied by the co-ordinates A, Jc ; therefore 

hx\ + Icy, = r 2 , (1) 

hx 2 -f- Jcy 2 = r 2 . (2) 

From equations (1) and (2) we see that the two points 
(#i> Vi)i (^2, 2/2) both satisfy the equation 

hx + Jcy = r 2 , (3) 

which, as its form shows, represents some straight line. There- 
fore equation (3) is the equation of the straight line passing 
through (# 1} y x ) and (x 2 , y 2 ) ; in other Avords, the equation 
required. 

The line represented by equation (3) is a real line, whether 
(/i, 1c) be outside or inside the circle. 

If the point (A, Jc) be outside the circle, this line is called 
the Chord of Contact of the two real tangents drawn from (A, Jc). 

If the point (A, Jc) be inside the circle, the points of contact 
and the tangents are imaginary, and we have a real line 
joining two imaginary points. 

79. The straight line passing through the points of contact 
of the tangents (real or imaginary) which can be drawn from 
any point to a circle is called the Polar of that point with 
respect to the circle ; and the point is called the Pole of that 
straight line with respect to the circle. 



98 ANALYTIC GEOMETRY. 

80. If the polar of a point P pass through Q, then the 
polar of Q will pass through P. 

Let P be the point (A, 1c), Q the point (A', £'), and let the 
equation of the circle be x 2 -\-y 2 — r 2 . 

Then the equations of the polars of P and Q are 

hx+hj=r\ (1) 

h'x.+ Vy=f*. (2) 

If Q be on the polar of P, its co-ordinates must satisfy 
equation (1) ; therefore 

AA'+M'-r 2 . 

But this is also the condition that P shall be on the line 
represented by (2) ; that is, on the polar of Q. Therefore P 
is on the polar of Q. 

81. If a straight line revolve about a fixed point Q, and P 
is the pole of that line, the locus of P is the polar of Q. 

For, since Q is on the polar of P, the point P must always 
be on the polar of Q (§ 80). 

82. If a point Q move along a fixed straight line, and P is 
the pole of that line, then the polar of Q will revolve about P. 

For, by hypothesis, the polar of P passes through Q (§ 80). 

83. The polar of a point with respect to a circle is perpen- 
dicular to the line joining the point to the centre of the circle. 

Let the equation of the circle be 

x2 + y 1 — r2 > 

and let P be any point (A, h). Then the equation of the 
polar of Pis hx + hj = r 2 . (1) 

And the equation of the line joining P to the centre of 
the circle is kx-hy = 0. (2) 



THE CIRCLE. 



99 



The form of equations (1) and (2) shows that the lines 
which they represent are perpendicular (§ 51). 

Figs. 31 and 32 illustrate the relations of poles and polars 
which have been established in §§ 80-83. 





100 



ANALYTIC GEOMETRY. 



84, To find a geometrical construction for the polar of a 
point with respect to a circle. 





Fig. 33. 



Fig. 34. 



If the notation of § 83 be retained, and OQ (Figs. 33 and 34) 
be the perpendicular from to the polar -of P, then (§ 55) 

OQ = 



VA 1 



Also OP-V/r + Z; 2 . 

Therefore ■_ OPx OQ = r\ 

Hence we have the following construction : 
Join OP, and let it cut the circle in A ; take Q in the line 
OP, such that 0P : OA=--OA: OQ, 

and draw through Q a line perpendicular to OP. 

85. To find the length of the tangent drawn from any point 
(A, k) to the circle r x _ a y _|_ ( y _ iy _ ^ = q. (1) 

Let P (Fig. 35) be the point (h, Jc), Q the point of contact, 
Cthe centre of the circle ; then, since PQC is a right angle, 

PQ 2 = PC 2 - QC\ 

Now PC 2 = (h - a) 2 + (& - b) 2 , QC 2 = r\ 

Therefore PQ 2 = (h- a) 2 + (Jc - b) 2 - r 2 . 



THE CIECLE. 



101 



Hence PQ 2 is found, by simply substituting the co-ordinates 
of P in the left-hand member of equation (1). 

If for brevity we write S instead of (x — a) 2 -f- (y — b) 2 — ?* 2 , 
then the equation S= will represent the general equation 
of the circle after division by the common coefficient of x 2 
and y 2 , and we may state the above result as follows : 

Pf S=0 be the equation of a circle, and the co-ordinates of 
any point be substituted for x and y in S, the result will be 
equal to the square of the length of the tangent drawn from the 
point to the circle. 

If the point is inside the circle, the square is negative, and 
the length of the tangent imaginary. 

86. Two circles intersect each other ; to find the equation of 
the straight line passing through the points of intersection. 





Fig. 35. 



Fig. 36. 



Let the equations of the circles be 

(*-a) 2 +G/-5)'-r'=0, (1) 

( x - a y + (y-by-r" = 0. (2) 

Subtract one of these equations from the other; then 
2(a-a')x + 2(b-bjy = a 2 -a ,2 +b 2 -b ,2 -(r 2 -r ,2 ) = 0. 

This is the equation required ; for it is the equation of 
some straight line, and its locus passes through the intersec- 
tions of the loci represented by (1) and (2) (§ 59). 



102 ANALYTIC GEOMETRY. 

If we write /Si and S 2 for the left-hand members of equations 
(1) and (2) respectively, the result may be thus stated : 

If Si = 0, S 2 = be the equations of two circles, then will 
the equation /Si — S 2 = 0, or /Si = S 2 be the equation of the 
straight line through their points of intersection. 

Although the two circles S x = 0, S 2 = may not cut each 
other in real points, the straight line Si = S 2 will always be 
real, provided the constants in it are real; so that we have 
a real straight line passing through imaginary points. 

But another meaning may be given to the equation S L — S 2 . 

For if (x, y) denote any point in the line /Si = S 2 , then /Si 
is equal to the square of the tangent from (x, y) to the circle 
Si = 0, and S 2 is equal to the square of the tangent from 
(x, y) to the circle S 2 = (§ 85). 

Hence the tangents drawn to the two circles from any 
point in the straight line /Si = S 2 are equal. 

The straight line /Si = S 2 is called the Eadical Axis of the 
two circles /Si — 0, S 2 — 0. 

It may be defined either as the straight line passing through 
the points of intersection (real or imaginary) of the two circles, 
or as the locus of the points from which tangents drawn to the 
two circles are equal. 

87. The three radical axes of three circles, taken in pairs, 
'meet in a point 

Let #=0, Si ~ 0, #2 = be the equations of the circles, 
when the coefficient of x 2 in each is unity. 

Then the equations of their radical axes, taken in pairs, are 

S-Si = 0, Si-S 2 = y S-S 2 = 0. 

The values of x and y that will satisfy any two of these 
equations will also satisfy the third. Therefore the third 
axis passes through the point of intersection of the other 
two axes. The point of intersection of the three radical axes 
is called the Eadical Centre of the three circles. 



THE CIKCLE. 103 

Ex. 24. 

1. What is the equation of the diameter of the circle 
# 2 -j-y 2 — 20 which bisects chords parallel to the line 6x-\-7y 
+ 8 = 0? 

2. What is the equation of the diameter of the circle which 
bisects all chords whose inclination to the axis of x is 135° ? 

3. Prove that the tangents at the extremities of a diameter 
are parallel. 

4. Write the equations of the chords of contact in the circle 
% 2 + y 1 — ^ f° r tangents drawn from the following points : 
(r,r), (2r, 3r), (a + b, a-b). 

5. From the point (13, 2) tangents are drawn to the circle 
# 2 + y 2 — 49 ; what is the equations of the chord of contact ? 

6. What line is represented by the equation lix -f- ley = r 2 
when (A, k) is in the circle ? 

7. Write the equations of the polars of the following points 
with respect to the circle x 2 + y 2 = 4 : 

(i.) (2,3). (ii.) (3,-1). (iii.) (1,-1). 

8. Find the poles of the following lines with respect to the 
circle x 2 + y 2 = 35 : 

(i.) 4^ + 6y=--7. (ii.) 3x — 2y=5. (iii.) ax + by = l. 

9. Find the pole of Zx-^^y = 7 with respect to the circle 
x 2 + y 2 = U. 

10. Find the pole of Ax -{-By -f- C— with respect to the 
circle x 2 -f- y 2 = r 2 , 

11. Find the co-ordinates of the points where the line x — 4 
cuts the circle x 2 -f y 2 = 4 ; also find the equations of the tan- 
gents at those points, and show that they intersect in the 
point (1, 0). 



104 ANALYTIC GEOMETRY. 

12. If the polars of two points P, Q meet in JR, then E is 
the pole of the line PQ. 

13. If the polar of (A, 1c) with respect to the circle x 2 +y* = r 2 
touch the circle x 2 + y 1 = 2rx, then 1c 2 — 2rh = r 2 . 

14. If the pole lie in the circle x 2 + y 2 = 4 c 2 , then the 
polar will touch the circle 4(# 2 -f y 2 ) = <? 2 . 

15. Find the polar of the centre of the circle x 2 + if = r 2 . 
Trace the changes in the position of the polar as the pole is 
supposed to move from the centre to an infinite distance. 

16. What is the square of the tangent drawn from the 
point (h, 1c) to the circle x 2 + y 2 — r 2 ? 

17. Find the length of the tangent drawn from (2, 5) to 
the circle x 2 -f y 2 — 2 x — 3 y — 1 = 0. 

Find the radical axis of the circles 

18. (^ + 5) 2 + (y + 6) 2 = 9, (*-7)' + (y-ll)'=16. 

19. x 2 + y 2 + 2x + 3y-7=0, x 2 + if - 2x -y + 1=0. 

20. x 2 + y 2 + bx + by — c = 0, ax 2 + ay 2 + a 2 x + b 2 y = 0. 

21. Find the radical axis and length of the common chord 
of the circles 

x 2 _j_ yl _|__ ax _j_ -foy _j_ c _ Q^ x 2 _I_ y 2 _}_ S^; + 6^3/ + <? = 0. 

22. Find the radical centre of the three circles 

x 2 + y 2 + 4x + 7=0, 

2x 2 + 2y 2 + ?>x + 5y + 9 = 0, 

& + y 2 + y = °- 

23. Proye that the radical axis of two circles is perpen- 
dicular to the straight line joining their centres. 

24. Find a geometric construction for the radical axis of 
two circles which do not meet each other (see § 87 and Ex. 23). 



CHAPTER IV. 

DIFFERENT SYSTEMS OF CO-ORDINATES. 

Oblique Co-ordinates. 

88. When we define the position of a point or a line by 
reference to some system of points or lines regarded as fixed 
in position, we are said to employ a System of Co-ordinates. 

The system of co-ordinates which we have thus far em- 
ployed is called the Bectangnlar System, because the two fixed 
lines of reference are perpendicular to each other. It is the 
system to be preferred for most purposes on account of its 
simplicity. 

There are, however, two other systems in use, of such im- 
portance that we shall briefly describe and illustrate them. 



The first of these systems differs from the rectangular 
system simply in the fact that the axes of reference are not 
perpendicular to each other. 

Let OX, OF (Fig. 37) be two axes making an acute angle, 
XO Y= co, with each other. The position of the point P is 



106 ANALYTIC GEOMETRY. 

determined by stating its distance from each axis, measured 
along a line parallel to the other axis. 

If we draw PN\\ to OX, and PM || to OY, then the co- 
ordinates of P are 

NP=OM=x, MP = y. 



This system of co-ordinates is known as the Oblique System. 

Kectangnlar and oblique co-ordinates are called Parallel 
Co-ordinates ; also Cartesian Co-ordinates (from Descartes, who 
first used them). 

89. To find the equation of the straight line A C, referred to 
the oblique axes OX, Y (Fig. 38), having given the intercept 
OB — b and the angle XAC= y. 

Let P be any point (x, y) of the. line. Draw BB \\ to OX, 
meeting PM in B. Then, by Trigonometry, 

PB sin y y — b sin y 

. — _ 1 ,, or iL = 1 

BB sin (w — y) x sin (a> — y) 

If now we put m = - — - — 2 — W e obtain as the result an 
sin (a) — y) 

equation of the same form as [6], p. 38, 

y = mx + b. 

What does the value of m become in this equation when 

o) = 90° ? 



DIFFERENT SYSTEMS OF CO-ORDINATES. 



107 



90. Oblique co-ordinates are seldom used, because they 
generally lead to more complex formulas than rectangular 
ones. In certain cases, however, they may be employed to 
advantage. An example of this kind is furnished by problem 
No. 23, p. 71 : 

To prove that the medians of a triangle meet in one point. 

If a, b, c represent the three sides of the triangle, and we 
take as axes the sides a and b, then the equations of the sides 
and also of the medians may be written down with great ease, 
as follows : 



0. 



On comparing the equations of the medians, w T e see that if 
we subtract the second equation from the first, we obtain the 
third ; therefore the three medians must pass through the. 
same point (§ 59). 



The sides, 


y = o. 


x = Q, 




= 1. 




The medians, 












a b 


-i-o, 


x 2 y 


-1=0, 


X 

a 


y 

b 




Fig, 39. 



Polar Co-ordinates. 

91. There is another system of co-ordinates, called the Polar 
tern, which is often useful. 
Let O (Fig. 39) be a fixed point, AOA 1 a fixed straight 
line, P any point. Join OP. 



108 



ANALYTIC GEOMETRY. 



It is evident that we know the position of P, provided 
we know the distance OP and the angle which OP forms 
with OA. 

Thus, if w r e denote the distance OP by p, and the angle 
POA by 0, the position of P is determined if p and are 
known. 

p and are called the Polar Co-ordinates of P ; is called 
the Pole ; OA, the Polar Axis ; OP, the Kadius Vector of P. 




Fig. 40. 



Every point in a plane is perfectly determined by a posi- 
tive value of p between and oo, and a positive value of 
between 0° and 360° (or and 2tt, circular measure). But 
in order to be able to represent by a single equation all the 
points of a geometric locus, it is necessary to admit negative 
values of p and 6, and to adopt conventions suitable for this 
purpose. 

It is agreed that 6 shall be considered positive when it is 
measured from the initial line, in the opposite direction to that 
of the motion of the hands of a watch ; and negative when 
measured in the same direction as this motion. 

It is also agreed that p, or OP, shall be considered positive 
when it forms one side of the angle 6, and negative when it 
does not. 



DIFFERENT SYSTEMS OF CO-ORDINATES. 



109 



For example, suppose that the straight line POP Y bisects 
the first and third quadrants, and that in this line we take 
points P, Pi, at the same distance OP= p from ; then 



P is the point (p, ^tt) or (— p, f it) or (— p, 



r), 



P x is the point (p, -f 7r) or (— p, ^tt) or ( p, — f 7r), etc. 





Fig. 41. 



Fig. 42. 



92. To find the polar equation of the circle. 

(i.) Let the pole be at the centre. Then, if r denote the 
radius, the polar equation is simply p = r. 

(ii.) Let the pole be on the circumference (Fig. 41), and 
let the diameter OB make an angle a with the initial line OA. 
Let P be any point (r, 6) of the circle. Join BJP. 

Then OP = OB cos POP, 

or P = 2rcos (0 — a). [23] 

If OP is taken as the initial line, the equation becomes 

P = 2 r cos 0. [24] 

(iii.) Let the pole be anywhere, and the centre the point 
(Jc, a). Then in the triangle OPC (Fig. 42) 

OP 2 - 2 OP x OCX cos POO+ OO 2 - CP 2 = 0, 
or P 2 — 2 9 k cos (0 - a) + ^ - r 2 = O, [25] 

the most general form of the polar equation of a circle. 



110 ANALYTIC GEOMETRY. 



Ex. 25. 



1. Find the distances from the point P in Fig. 38 to the 
two axes. 

2. Prove that the equation of a straight line, referred to 
oblique axes in terms of its intercepts, is identical in form 
with [7], p. 39. 

3. If the straight line P 2 OP 3 (Fig. 39) bisects the second 
and fourth quadrants, what are the polar co-ordinates of the 
points _P 2 and P 3 ? Give more than one set of values in each 
case. 

4. Construct the following points (on paper, take a = 1 in.) : 

*.°) («•!} («.-f} (-«>!) (-°'-i 

2a, |), (2a,.), (a cos 1 1), (a, £= ), (3a, |* 
-3o, — \ (4a, tan" 1 ^), Ala, tan" 1 - 

4- . . 

Note. The expression tan -1 - in higher Mathematics means " the 

4 3 

angle whose tangent is -•' 

5. If pi, p 2 denote the two values of p in equation [25], 
p. 109, prove that p x and p 2 = Z: 2 — r 2 . What theorem of Ele- 
mentary Geometry is expressed by this equation (i.) when 
the pole is outside the circle ? (ii.) when the pole is inside 
the circle ? 

6. Through a fixed point Pin a circle a chord AB is drawn, 
and then revolved about _P; find the locus of its middle point. 

Note. In such problems as this there is a great advantage in using 
polar equations. 

7. If p denote the distance from the pole to a straight line, 
a the angle between p and the polar axis, prove that the polar 
equation of the line is p cos (6 — a) =p. 



DIFFERENT SYSTEMS OF CO-ORDINATES. 



Ill 



Transformation of Co-ordinates. 

93. The equation of the same curve varies greatly in form 
and simplicity, according to the system of co-ordinates adopted, 
and the position of the fixed points and lines with respect to 
the curve. Hence it is sometimes useful to be able to deduce 
from the equation of a curve referred to one system of co- 
ordinates its equation referred to another system. This pro- 
cess is known as the Transformation of Co-ordinates. 

It consists in expressing the old co-ordinates in terms of 
the new, and then replacing in the equation of the curve the 
old co-ordinates by their values in terms of the new ; we thus 
obtain a constant relation between the new co-ordinates, which 
will represent the curve referred to the new axes. 

94. To change the origin to the point (/i, Jc) without changing 
the direction of the axes. 





Y 




Y 1 








a 


| X 









1 


M 


X 



Fig. 43. 



Let OX, OY be the old axes, O'X', O'Y' the new; and 
let (x, y), (V, y') he the co-ordinates of the same point P, 
referred to the old and new systems respectively. 

Then (Fig. 43) 



OA = h, AO'=h, OM=x, MP=y, 0'M' = x', M'P- 
x = OA + AM = OA + O'M' = x\A-K. 
y = MJP + M'P =AO'+ M'P =y' + L 



■y'. 



112 



ANALYTIC GEOMETRY. 



These relations are equally true for rectangular and oblique 
co-ordinates. 

Hence, to find what the equation of a curve becomes when 
the origin is transferred to a point (A, h), the new axes run- 
ning parallel to the old, we must substitute for x and y the 
values given above. 

After the substitution, we may, of course, write x and y 
instead of x ! and y f ; so that practically the change is effected 
by simply writing x-\- h in place of x, y-\-h in place of y. 

If, however, we wish to transform a point (x, y) from the 
old to the new system, we must write x — h in place of x and 
y — k in place of y. 

95. To change the equation of a curve from one rectangular 
system to another, the origin remaining the same. 




Let (x, y) be a point P referred to the old axes OX, Y\ 
(x',y'), the same point referred to the new axes OX\ OY f 
(Fig. 44). Then 

OM=x, MP = y t ON=x\ NP = y\ 

Let the angle XOX'=6. Draw XQ, NE JL to PM, OX, 

respectively ; then 

NPQ = QJSTO =■- NOP = 6. 



DIFFERENT SYSTEMS OF CO-ORDINATES. 113 

Hence OM= OB -BM= OE — NQ= OJVcos 6 - iWsin 0. 

Or x — x 1 cos — y f sin 6. 

And PM= MQ+ QP--=RN+ QP= OiVsin + PJV cos 6. 

Or y — # f sin + y' cos 0. 

Therefore, to find what the equation of a curve becomes 
when referred to the new axes, we must write 

x cos — y sin 6 for x, x sin + y cos . for-?/. 

96. To transform an equation from one rectangular system 
to another, both the origin and the direction of the axes being 
changed. 

First transform the equation to axes through the new origin, 
parallel to the old axes. Then turn these axes through the 
required angle. 

If (A, lc) is the new origin referred to the old axes, 6 the 
angle between the old and new axes of %, we obtain as the 
values of x and y for any point P, in terms of the new co- 
ordinates, t , 1 A f ' A 

x = h + % cos 6 — y sin 0, 
y — k + x' sin + y f cos 0. 

In making all these transformations, attention must be paid 
to the signs of h, k, and 6. 

97. :7b transform an equation from rectangular to oblique 
axes, the origin remaining the same. 

Let a, (3 be the angles formed by the positive directions of 
the new axes OX\ Y' (Fig. 45) with the positive direction of 
OX. Let the old co-ordinates of a point P be x, y ; and the 
new co-ordinates, x\ y'. Then from the right triangles OBX, 
PQJYwe readily obtain the formulas 

x = x* cos a -f- y 1 cos /?, 
y — x f sin a -f- y 1 sin f3. 

Investigate the special case when (3 = a -f- 90°. 



114 



ANALYTIC GEOMETRY. 



98. To change an equation from polar to rectangular co- 
ordinates. 




R X 




M X 



Fig. 46. 



Let the co-ordinates of a point JP be x, y referred to the 
rectangular system, and p, referred to the polar system. 

(i.) Let the origin of rectangular co-ordinates be the pole, 
and let the polar axis coincide with the axis of x. 
Then (Fig. 46) OM==OPco , FOMt 

PM=OPsm f POM. 
Or x = p cos 0, 

y = p sin 0. 

(ii.) If the pole is the point (h, k), we have 

x = A + p cos } 
y — 1c + p sin 6. 

(iii.) If the pole coincides with the origin, but the polar 
axis OA makes the angle a with the axes of x, we obtain 
X = p COS (0 + a), 
y = psin (0 + a). 

(iv.) If the pole is the point (A, k), and the polar axis 
makes the angle a with the axis of x, 

x = h + p cos (0 -f- a), 
y = Jc -f p sin (0 + a). 



DIFFERENT SYSTEMS OF CO-ORDINATES. 115 

99. To change an equation from polar to rectangular co- 
ordinates. 

From the results in cases (i.) and (ii.) of § 98 (the only 
cases of importance), we readily obtain 

In case (i.), r 2 = x 2 + y 2 , tan0=-- 

In case (ii.), r 2 = (x - h) 2 + (y — h)\ tan = £■ 

100. The degree of an equation is not altered by any alteration 
of the axes. 

For, however the axes may be changed, the new equation 

is always obtained by substituting for x and y expressions of 

the form . 7 , 1 , . t, , • 

ax + by + c and a'x + 6 f y -[- c . 

These expressions are of the first degree, and therefore, if 
they replace x and y in the equation, the degree of the equa- 
tion cannot be raised. Neither can it be lowered ; for if it 
could be lowered, it might be raised by returning to the 
original axes, and therefore to the original equation. 

Ex. 26. 

1. What does the equation y 2 — 4^ + 4y + S = become 
when the origin is changed to the point (1, — 2) ? 

Transform the equation of the circle (x — a) 2 + (y — b) 2 = r 2 
by changing the origin 

2. To the centre of the circle. 

3. To the left-hand end of the horizontal diameter. 

4. To the upper end of the vertical diameter. 

5. What does the equation x 2 -\-y 2 — r 2 become if the axes 
are turned through the angle a ? 

6. What does the equation x 2 — y 2 = a 2 become if the axes 
are turned through — 45° ? 



116 ANALYTIC GEOMETRY. 

7. The equation of a curve referred to rectangular axes is 
x — xy — y = 0. Transform it to a new system, whose origin 
is the point (—1, 1), and whose axes bisect the angles formed 
by the old axes. 

8. Change the following equations to polar co-ordinates, 
taking the pole at the origin and the polar axis to coincide 
with the axis of x : 

(i.) x 2 + y 2 = & 2 . (ii.) x 2 — y 2 = a 2 . 

9. Change the equation # 2 =4aa; to polar co-ordinates, 
(i.) taking the pole at the origin ; (ii.) taking the pole, at 
the point (a, 0). 

10. Change the following equations to rectangular co-ordi- 
nates, the origin coinciding with the pole, and the polar axis 
with the axis of x : 

(i.) p = a, (ii.) p — a cos 8, (iii.) p 2 cos 26 = a 2 . 

Transform the following equations by changing the origin 
to the point given as a new origin : 

11. # + y + 2 = ; the new origin (— 2, 0). 

12. 2x — 5y — 10 = 0; the new origin (5, —2). 

13. 3x 2j r4:xy-{-y 2 — 5x — 6y — 3 = 0; new origin (J, —4). 

14. x 2 -\-y 2 — 2x — 4y = 20 ; new origin (1, 2). 

15. x 2 — 6xy -{- y 2 — 6x J r2y-\-l= ; new origin (0, —1). 

16. Transform the equation x 2 — y 2 -{- 6 = by turning the 
axes through 45°. 

17. Transform the equation (x + y — 2 a) 2 = 4:xy by turn- 
ing the axes through 45°. 

18. Transform the equation 9x 2 — 16y 2 = 144 to oblique 
axes, such that the new axis of x makes with the old axis 
of x a negative angle whose tangent = — | ; and the new axis 
of y makes with the old axis of x a positive angle whose tan- 
gent is f . 



DIFFERENT SYSTEMS OF CO-ORDINATES. 117 

Ex. 27. (Review.) 

1. Find the distance from the point (—2b, b) to the origin, 
the axes making the angle 60°. 

2. The axes making the angle w, find the distance from the 
point (1, —1) to the point (—1, 1). 

3. The axes making the angle co, find the distance from the 
point (0, 2) to the point (3, 0). 

Determine the distance between the following points referred 
to polar co-ordinates : 

4. (a, 6) and (5, 0). 

5. (a, 6) and (a, — 0). 

6. (a, 6) and (—a, -6). 

7. (2 a, 30°) and (a, 60°). 

8. Show that the polar co-ordinates (p, 0), (— p, 7r + 0), 
(— p, 6 — tt) all represent the same point. 

9. Transform the equation 8^ 2 + &ry+4y 2 +12^ + 8?/ +1 = 
to the new origin (—-£, —2"). 

10. Transform the equation 6x 2 -\-3y 2 — 24:X J r 6 = to the 
new origin (2, 0). 

x 1/ 

11. Transform the equation --f-~ = l by changing the 

origin to the point [-' -j and turning the axes through an 
angle <£, such that tan </> = 

12. Transform the equation 17a; 2 -16xi/ +17 y 2 = 225 to 
axes which bisect the axes of the old system. 

Transform the following rectangular equations to polar 
equations, the polar axis in each case coinciding with the 
axis of x, and the pole being at the point whose co-ordinates 
are given : 



118 ANALYTIC GEOMETRY. 

13. a? + 2f = 8ax] the pole (0, 0). 

14. x 2 + y 2 = Sax ; the pole (4a, 0). 

15. y 2 -6y-5x + 9 = 0] the pole (f, 3). 

16. x 2 - y 2 - 4.x - 6y - 54 = ; the pole (2, - 3). 

17. (x 2 + y 2 ) 2 = JP(a? - y 2 ) ; the pole (0, 0). 

Transform the following polar equations to rectangular axes, 
the origin being at the pole and the axis of x coinciding with 
the polar axis : 

18. p 2 sin 20 = 2a*. 

19. p = ks'm20. 

20. p(sin30 + cos30) = 5£sin0cos0. 

21. Through what angle must the axes of a rectangular 
system be turned in order that the new axis of x may pass 
through the point (5, 7) ? 

22. The equation of a straight line in rectangular axes is 
Ax-\-By-\-C= 0. Through what angle must the axes be 
turned in order 

(i.) that the term containing x may disappear ? 
(ii.) that the term containing y may disappear ? 

23. Deduce the following formulas for changing from one 
oblique system to another, the origin remaining the same : 

x 1 sin (to — a) , ?/ sin (to — (3) 

sin (o sin to 

x f sin a . v f sin 6 
sm to sin to 

Note. In these formulas a> denotes the angle formed by the old axes, 
a and £ those formed by the positive directions of the new axes with the 
positive direction of the old axis of x. 

24. From the formulas of No. 23 deduce those of § 97. 



CHAPTER V. 

THE PARABOLA. 

The Equation of the Parabola. 

101. A Parabola is the locus of a point whose distance from 
a fixed point is always equal to its distance from a fixed 
straight line. 

The fixed point is called the Focus ; the fixed straight line, 
the Directrix. 

The straight line which passes through the focus, and is per- 
pendicular to the directrix, is called the Axis of the parabola. 

The intersection of the axis and the directrix is called the 
Foot of the axis. 

The point in the axis half way between the focus and the 
directrix is, from the definition, a point of the curve ; this 
point is called the Yertex of the parabola. 

The straight line joining any point of the curve to the focus 
is called the Focal Radius of the point. 

A straight line passing through the focus and limited by 
the curve is called a Focal Chord. 

The focal chord perpendicular to the axis is called the Latus 
Rectum or Parameter. 

102. To construct a parabola, having given the focus and 
the directrix. 

I. By Points. Let F (Fig. 47) be the focus, CE the direc- 
trix. Draw the axis FD, and bisect FD in A ; then A is the 
vertex of the curve. At any point M in the axis erect a per- 
pendicular. From F as centre, w T ith DM as radius, cut this 



120 



ANALYTIC GEOMETRY. 



perpendicular in P and Q ; then P and Q are two points of 
the curve, for FP = P3I— distance of P or Q from CE. In 
the same way we can find as many points of the curve as we 
please. After a sufficient number of points has been found, 
we draw a smooth curve through them. 





Fig. 47. 



Fig. 48. 



II. By Motion. Place a ruler so that one of its edges 
shall coincide with the directrix DE (Fig. 48). Then place 
a triangular ruler BCE with the edge CE against the edge 
of the first ruler. Take a string whose length is equal to 
PC; fasten one end at B and the other end at F. Then 
slide the ruler BCE along the directrix, keeping the string 
tightly pressed against the ruler by the point of a pencil P. 
The point P will trace a parabola ; for during the motion we 
always have PF= PC. 

103. To find the equation of the parabola, when its axis is 
taken as the axis of x and its vertex as the origin. 

Let F (Fig. 49) be the focus, CE the directrix, DFX the 
axis, A the vertex and origin ; also let 2p denote the known 
distance FD. 

Let P be any point of the curve ; then its co-ordinates are 

AM=x, PM=y. 



THE PARABOLA. 



121 



Draw PC J. to CE; then by the definition of the curve 
FP = PO= DM. 

,2 



Therefore 
Now 
and 



Fl * = -DM . 

FF = PM 2 + FM 2 = y 2 + (x -p)\ 

[26] 



DM = (x+p) 2 . 
Therefore y 2 + (x — p) 2 — (x +p) 2 . 
Whence V 2 = 4=px. 

This is called the principal equation of a parabola. 





Y 




Q 


V 






p 


D 
E 


A 


i * 


if x 



104, Since y 2 and /> in equation [26] are positive, x must 
always be positive ; therefore the curve lies wholly on the 
positive side of the axis of y. 

A further examination of equation [26] shows that the 
curve, (i.) passes through the origin, (ii.) is symmetrical with 
respect to the axis of x, (iii.) extends towards the right with- 
out limit, and (iv.) recedes from the axis of x without limit. 

105. Any point (h, Jc) is outside, on, or inside the parabola 
y 2 = 4p#, according as Jc 2 — 4cph is positive, zero, or negative. 

• Let Q be the point (h, Jc), and let its ordinate meet the 
curve in JP. 



122 ANALYTIC GEOMETRY. 

If h 2 — 4pA — 0, the point (A, k) satisfies equation [26], and 
therefore Q coincides with P. 

If k 2 — 4:ph is positive, or h 2 > 4pA, then, since P3f 2 = 4:ph r 
we have QM* > PM\ or QM>PM] hence Q is outside 
the curve. 

If k 2 — 4:ph is negative, we may prove similarly that Q 
must be inside the curve. 

106. To find the lotus rectum of a parabola. 

The common abscissa of the two points where the latus 
rectum meets the curve =p. Substituting this value for x 
in equation [26], we have y = zh 2 p. Therefore the latus 
rectum = 4p. 

107 +^To find the points in which the straight line y=mx-\-c 
meets the parabola y' 2 — 4:px. 

The co-ordinates of these points must satisfy both equations ; 
hence, at a common point, we have the relation 

f-^pir^A- (i) 

\ m J 

Since (1) is a quadratic equation, we see that every straight 
line meets a parabola in two points. Solving (1) ; we obtain 
for the ordinates of these two points 

2p ± 2p Jp-mc (2) 

a m m \ p v J 

whence it appears that the points are real, coincident, or 
imaginary, according &s p — mc is positive, zero, or negative. 



108. To find the equation of a parabola whose axis is par- 
allel to the axis of x. 

Let the vertex be the point (a, b), and let 2p — distance 
from focus to directrix. Then the focus will be the point 
(a -\-p, 6), and the directrix will be the line x = a —p. 



THE PARABOLA. 123 

The distance of any point (x, y) from the focus is 



-V(x-a-p) 2 +(y-h) 2 , 

and its distance from the directrix is 

x — a-\-p. 

If (x, y) is a point of the parabola, these distances are equal ; 
putting them equal, and reducing, we obtain 

y 2 - ±px — 2by+b 2 -4.ap = 0. (1) 

Hence we may infer that in general an equation having 

the form ^ + Ax + By+C=Q (2) 

represents a parabola having its axis parallel to the axis of x. 
By equating coefficients in (1) and (2), we obtain 

, A B 2 -±C , B 

whence the following results easily follow : 
The latus rectum = — A. 

The vertex is the point ( > — — 

^ \ 4.A 2 

The focus is the point 



4:A 4 

The axis is the line y = — ■ — 

y 2 

The directrix is the line x — — — 

AA 

If A is negative, the parabola lies to the right of the axis 
of y, and may be called right-handed. 

If A is positive, the parabola lies to the left of the axis of y, 
and may be called left-handed. 



124 ANALYTIC GEOMETRY. 



"Ex. 28. 



1. Show that the distance of any point of the parabola 
y 2 — ipx from the focus is equal to p + x. 

2. Find the equation of a parabola, taking as axes the 
axis of the curve and the directrix. 

3. Find the equation of a parabola, taking the axis of the 
curve as the axis of x and the focus as the origin. 

4. The distance from the focus of a parabola to the direc- 
trix — 5. Write its equation, 

(i.) If the origin is taken at the vertex, 
(ii.) If the origin is taken at the focus, 
(iii.) If the axis and directrix are taken as axes. 

5. The distance from the focus to the vertex of a parabola 
is 4. Write its equations for the three cases enumerated in 

No. 4. 

6. For what point of the parabola y 2 = 18# is the ordi- 
nate equal to three times the abscissa ? 

7. Find the latus rectum for the following parabolas : 

y 2 = 6#, y 2 = 15 #, by 2 = ax. 

Find the points common to the following parabolas and 
straight lines : 

8. y 2 = 9x, 3.z — 7y + 30 = 0. 

9. y 2 ^Sx, x-^y + 12-0. 

10. y 2 =Ax, x = 9, x = 0, x = — 2. 

11. y 2 = 4:x, y = 6 7 y— ■ — 8. 

12. What must be the value of p in order that the parabola 
y 2 = 4:px may pass through the point (9, —12) ? 



THE PARABOLA. 125 

13. For what point of the parabola y 2 = 32# is the ordi- 
nate equal to 4 times the abscissa ? 

14. The equation of a parabola is y 2 = Sx. What is the 
equation of (i.) its axis, (ii.) its directrix, (iii.) its latus rectum, 
(iv.) a focal chord through the point whose abscissa = 8, (v.) a 
chord passing through the vertex and the negative end of the 
latus rectum ? 

15. The equation of a parabola is y 2 = 16x. Find the 
equation of (i.) a chord through the points whose abscissas 
are 4 and 9, and ordinates positive ; (ii.) the circle passing 
through the vertex and the ends of the latus rectum. 

16. If the distance of a point from the focus of the parabola 
y 2 = ^.px is equal to the latus rectum, what is the abscissa of 
the point ? 

17. In the parabola y 2 = 4:px an equilateral triangle is 
inscribed so that one vertex is at the origin. What is the 
length of one of its sides ? 

18. A double ordinate of a parabola = Sp. Prove that 
straight lines drawn from its ends to the vertex are perpen- 
dicular to each other. 

Explain how to construct a parabola, having given 

19. The directrix and the vertex. 

20. The focus and the vertex. 

21. The axis, vertex, and latus rectum. 

22. The axis, vertex, and a point of the curve. 

23. The axis, focus, and latus rectum. 

24. The axis, directrix, and one point. 

25. The axis and two points. 



126 ANALYTIC GEOMETRY. 

26. Determine, as regards size and position, the relations 
of the following parabolas : 

(i.) y 2 = 4 j £>s, (ii.) y 2 --—4:px, (iii.) x 2 = ipy, (iv.) x 2 = — ipy. 

27. What is the locus of the equation y 2 + Ax + By + C ' = 
in the following special cases : 

(i.) A = 0? (iii.) (7=0? (v.) .4 = C=0? 

(ii.) P = 0? iiY.yA=B = Q? (vi.) P = C=0? 

28. Show that in general the equation x 2j r Ax J r By-\- 0= 
represents a parabola whose axis is parallel to the axis of y ; 
and determine the latus rectum, the vertex, the focus, the 
axis, and the directrix. 

Find the latus rectum, vertex, focus, axis, and directrix of 
the following parabolas : 

29. y 2 - 12s + 84 = 0. 33. s 2 -12y + 84 = 0. 

30. y 2 - 12s -84 = 0. 34. y 2 - Sx- Sy + 64 = 0. 

31. y 2 + 12s + 84 = 0. 35. l+2s+3y 2 = 0. 

32. y 2 + 12s -84 = 0. 36. y = s 2 -s-2. 

37. y 2 -4s + 6y + l = 0. 

Tangents and Normals. 

109. To find the equation and the normal of the tangent to 
the parabola y 2 = 4^s, at the point of contact (x Xi y x ). 

If P, Q are the points (s 1? y Y ), (s 2 , y 2 ), the equation of a 
straight line through P and Q is 



S Sj X2 Sj 

If P and Q are points of the parabola y 2 = 4j2S, 

3/j 2 = 4^s x , 
y 2 2 = 4^s 2 . 



(1) 





THE 


PARABOLA. 


"Whencp 


3/2- 


yiL 


4j9 




x 2 


x\ 


y 2 + yi 


By substitution, 


equatior 


i(i) 


becomes 




y- 


yi_ 


. 4^ . 

> 



127 



x — x x y 2 + y x 

whence, by clearing of fractions, and remembering that 
yi = ipx, we obtain the equation of a secant in the form 

2/G/i + y 2 ) - 2A2/2 = ±px. (2) 




Fig. 50. 



Now let the secant turn about P till Q coincides with P ; 
then % 2 = x 1 , y 2 = yi, the secant becomes the tangent at P, 
and its equation reduces to 

ViV = 2p(x + oc 1 ). [27] 

m 

The normal passes through (x 1} y x ), and is perpendicular to 
the tangent ; hence (§ 50) its equation is 



%P(V — 2/j) + y^n — oc x ) = O. 



[28] 



128 ANALYTIC GEOMETRY. 

110. If we make y = in equations [27] and [28], we obtain 

x = — x x and x = x x -{- 2p. 

Hence the values of the subtangent MT and the subnormal 
MN are MT= 2 ^ MN= ^ 

Therefore 

(i.) The subtangent is bisected at the vertex. 
(ii.) The subnormal is constant, and equal to the distance 
from the focus to the directrix. 

111. In the triangle FPT(Fip;. 50) we have 

FT= FA + AT=p + x, 

FP = PC = DM=BA +AM=p + x. 

Therefore FT= FP. 

Hence the angle 

FPT= PTF= TPC, or 

The tangent to a parabola at any point makes equal angles 
with the focal radius and the line passing through the point 
parallel to the axis. 

112. To find the equation of a tangent to a parabola in terms 
of its slope. 

From the result obtained in § 107 we see that the straight 
line y — mx -f- c touches the parabola y 2 = 4:jjx, when 

mc —p, 

P 

or c = — - 

m 



Hence, for all values of m, the straight line 

y = mx - 
will touch the parabola y 2 = ipx. 



y — mx 4- — 
J m 



THE PARABOLA. 129 

Ex. 29. 

1. The normal to a parabola at any point bisects the angle 
between the focal radius and the line drawn through the 
point parallel to the axis. 

Note. The use of parabolic reflectors depends on this property. A 
ray of light issuing from the focus and falling on the reflector is reflected 
in a line parallel to the axis of the reflector. 

2. Explain how to draw a tangent and a normal to a given 
parabola at a given point. 

3. Prove that FC (Fig. 50) is perpendicular to FT. 

P 

4. Prove that the tangent y = mx + — touches the parabola 

f p 2p\ 
y 2 = 4px at the point ^— 2 > —J- 

5. Prove that the equation of a normal to the parabola 
y 2 = 4:px in terms of its slope is y = mx — wp(2 + ^ 2 )- 

6. What are the equations of a tangent and a normal to 
the parabola y 2 = 5x, passing through the point w T hose abscissa 
is 20 and ordinate positive ? 

7. What are the equations of the tangents and the normals 
to the parabola y 2 = 12x, drawn through the ends of the latus 
rectum ? Find the area of the figure which they enclose. 

8. Given the parabola y 2 = 10x. Through the point w r hose 
abscissa is 7 and ordinate positive a tangent and a normal are 
drawn. Find the lengths of the tangent, the normal, the 
subtangent, and the subnormal. 

9. A tangent to the parabola y 2 = 20x makes with the axis 
of x an angle of 45°. Determine the point of contact. 

10. Show that the focus i^(Fig. 50) is equidistant from the 
points F, T, iV". What easy w r ay of drawing a tangent and a 
normal is suggested by this theorem ? 



130 ANALYTIC GEOMETRY. 

11. If i^is the focus of a parabola, and Q, R denote the 
points in which a tangent cuts the directrix and the latus 
rectum produced, prove that FQ = FR. 

12. Prove that tangents drawn through the ends of the 
latus rectum are _L to each other. 



13. Find the distances of the vertex and the focus from the 

V 
tangent y = mx + — • 

14. The points of contact of two tangents are (x ly y^) and 
(x 2 , y 2 ). Find their point of intersection. 

15. A tangent to the parabola y 2 = ipx cuts equal inter- 
cepts on the axes. What is its equation ? What is the point 
of contact ? What is the value of the intercept ? 

16. Through what point in the axis of x must tangents to 
the parabola y 2 = £px be drawn in order that they may form 
with the tangent, through the vertex, an equilateral triangle ? 

17. For what point of the parabola y 2 — ipx is the normal 
equal to twice the subtangent? 

18. For what point of the parabola y 2 = 4:px is the normal 
equal to the difference between the subtangent and the sub- 
normal ? 

19. Find the equation of a tangent to the parabola y 2 = 5x 
parallel to the straight line 3# — 2"y-f-7=0. Also find the 
point of contact. 

20. Find the equation of the straight line which touches 
the parabola y 2 = 12x and makes an angle of 45° with the 
line y = 3x — 4. Also find the point of contact. 

21. Find the equation of a straight line which touches the 
parabola y 2 = 16x and passes through the point (—4, 8). 

22. If a normal to a parabola meet the curve again in the 
*>oint Q, find the length of FQ. 



THE PAEABOLA. 131 

23. Prove by the secant method that the equation of a tan- 
gent to the parabola y 2 == ^px — 4p 2 , at the point (x lt y^) is 

y<y ■■= 2p(x + x,) — 4p\ 

24. Find the equations of the tangents and normals to the 
parabola y 2 — 8x — 6y — 63 = 0, drawn through the points 
whose common abscissa ==. — 1. 

25. What are the equations of tangents to the following 
parabolas : 

(i.) y 2 = — ipx ? (ii.) x 2 = 4py ? (iii.) x 2 = — 4py ? 



Ex. 30. (Review.) 

Note. If not otherwise specified, the axis of the parabola and the 
tangent at the vertex are to be assumed as axes of co-ordinates. 

What is the equation of a parabola, 

1. If the axis and directrix are taken as axes, and the focus 
is the point (12, 0) ? 

2. If the axis and tangent at the vertex are the two axes, 
and (25, 20) is a point on the curve ? 

3. If the same axes are taken, and the focus is the point 
C-4i,0)? 

4. If the axis is taken as the axis of x, the vertex is the 
point (5, —3), and the latus rectum = 5J? 

5. If the axis is the line y = — 7, the abscissa of the vertex 
== 3, and one point is (4, — 5) ? 

6. If the curve passes through the points (0, 0), (3, 2), 
(3,-2)? 

7. If the curve passes through the points (0, 0), (3, 2), 
(-3,2)? 



132 ANALYTIC GEOMETRY. 

8. What is the latus rectum of the parabola 2y 2 — 3#? 
What is the equation of its directrix, and of the focal chords 
passing through the points whose abscissa = 6 ? 

9. Describe the change of form which the parabola y 2 =4:px 
undergoes as we suppose p to diminish without limit. 

10. Find the intercepts of the parabola y 2 -\-AiX — 6y— 16 = 0. 

11. One vertex of an equilateral triangle coincides with the 
focus, and the others lie in the parabola y 2 = 4:px. Find the 
length of one side. 

12. The latus rectum of a parabola = 8 ; find 

(i.) Equation of a tangent through its positive end. 
(ii.) Distance from the focus to this tangent, 
(iii.) Equation of the normal at this point. 

13. What is the equation of the chord passing through the 
two points of the parabola y 2 =Sx for which ^=2, y l > 0, 
and x 2 =18, y 2 < 0? 

14. Find the equation of the chord of the parabola y 2 = 4jp# 
which is bisected at a given point (x^ y x ). 

15. In what points does the line x J r y = 12 meet the para- 
bola y 2 + 2*-12y + 16 = 0? 

16. In what points does the line 3y = 2^ + 8 meet the 
parabola y 2 — ix — 8y + 24 = ? - 

17. Find the equations of tangents from the origin to the 
parabola (y — b) 2 = 4tp(x — a). 

18. Describe the position of the parabola y 2 -{- 2 x ~\- 4: -— 
with respect to the axes, and determine its latus rectum, 
vertex, focus, and directrix. 

19. What is the distance from the origin to a normal drawn 
through the end of the latus rectum of the parabola 

y 2 = 4 a(x — a)? 



THE PARABOLA. 133 

Find the equation of a parabola, 

20. If the equation of a tangent is 4y = Sx — 12. 

21. If a focal radius = 10, and its equation is 3?/ — 4 # — 8. 

22. If for a point of the curve the focal radius = r, the 
length of the tangent = t. 

23. If for a point of the curve the focal radius — r, the 
length of the normal = n. 

24. If for a point of the curve the length of the tangent = t, 
the length of the normal = n. 

25. If for a point of the curve the focal radius = r, the 
subtangent = s. 

26. Two parabolas have the same vertex, and the same 
latus rectum 4^, but their axes are J_ to each other. What 
is the length of their common chord ? 

27. Through the three points of the parabola y 2 = 12#, 
whose ordinates are 2, 3, 6, tangents are drawn. Show that 
the circle circumscribed about the triangle formed by the 
tangents passes through the focus. 

28. A tangent to the parabola y 1 = ^px makes the angle 
30° with the axis of x. At what point does it cut the axis ? 

29. For what point of the parabola y 1 = Apx is the length 
of the tangent equal to 4 times the abscissa of the point of 
contact ? 

30. The product of the tangent and normal is equal to 
twice the square of the ordinate of the point of contact. Find 
the point of contact and the inclination of the ordinate to the 
axis of x. 

31. Two tangents to a parabola are perpendicular to each 
other. Find the product of their subtangents. 



134 ANALYTIC GEOMETRY. 

32. Prove that the circle described on a focal radius as 
diameter touches the tangent drawn through the vertex. 

33. Prove that the circle described on a focal chord as 
diameter touches the directrix. 

Find the locus of the middle points 

34. Of all the ordinates of a parabola. 

35. Of all the focal radii. 

36. Of all the focal chords. 

37. Of all chords passing through the vertex. 

38. Of all chords that meet at the foot of the axis. 

Two tangents to the parabola y 2 = ipx make the angles 
6, ! with the axis of x ; find the locus of their intersection 

39. If cot 6 + cot 0' = h 41. If tan 6 tan ff = k 

40. If cot 6 - cot 6' = L 42. If sin 6 sin & = k 

43. Find the locus of the centre of a circle which passes 
through a given point and touches a given straight line. 

SUPPLEMENTARY PROPOSITIONS. 

113. Two tangents can be drawn to a parabola from any 
point ; and they will be real, coincident, or imaginary, accord- 
ing as the point is without, on, or within the curve. 

The tangent y = mx + — will pass through the point (h, h) 

if m 

1c = mh + — ; 
m 

that is, if hm 2 — lcm-\-p — §', 

whence m = l±^HM. 

2p 



THE PARABOLA. 135 

Since there are two values of m, two tangents can be drawn 
through the point (A, k). 

The values of m are real, coincident, or imaginary, accord- 
ing as k 2 — 4:hp is positive, zero, or negative; that is (§ 105), 
according as (A, k) is without, on, or within the curve. 

114. To find the equation of the straight line through the 
points of contact of the two tangents drawn to the parabola 
y 2 = 4:px from the point (A, k). 

If (xi, yi) and (x 2 , y 2 ) are the points of contact, the equa- 
tions of the tangents are 

y 1 y = 2p(x + x 1 ), 
y 2 y = 2p(x + x 2 ). 

Since (A, k) is in both these lines, 

ky l = 2p(x 1 + h), (1) 

ky 2 = 2p{x 2 + h). (2) 

But equations (1) and (2) are the conditions which make 
the points (x\, y^) and (# 2 , y 2 ) lie in the straight line whose 
equation is 

Hence (3) is the equation required. 

115. The straight line joining the points of contact of the 
two tangents (real or imaginary) from any point P to a para- 
bola is called the Polar of P with respect to the parabola ; &nd 
the point P is called the Pole of the straight line with respect 
to the parabola. 

The propositions in §§ 80-82, relating to poles and polars 

with respect to a circle, also hold true for poles and polars 

with respect to a parabola, and may be proved in exactly the 
same way. 



136 



ANALYTIC GEOMETRY. 



116. To find the locus of the middle points of parallel chords 
in the parabola y 2 = 4p#. 



\2>— ^C 




s/ / 


\ 




I / M 


7 


'A 





Fig. 51. 



Let the equation of any one of the chords PQ (Fig. 51) be 
y = mx -f- c, and let it meet the curve in the points (x u y 1 ) } 
O2, 3/2). 

Then (§109) m = -*£-. (1) 

Let (x, y) be its middle point M\ then 2y = y 1 + 3/ 2 . By 
substitution in (1) we obtain 

2p 



y = - 



(2) 



a relation which holds true for all the chords, because m is 
the same for all the chords. The required locus, therefore, 
is represented by (2), and is a straight line parallel to the 
axis of x. 

The locus of the middle points of a system of parallel chords 
in a parabola is called a Diameter \ and the chords are called 
the Ordinates of the diameter. 



THE PARABOLA. 137 

Therefore every diameter of a parabola is a straight line 
parallel to its axis. 

Conversely, every straight line parallel to the axis is a 

2p 
diameter ; for m, and therefore ■ — > may have any value what- 
ever. 

117. Let the diameter through M meet the curve at 8, and 
conceive the straight line PQ to move parallel to itself till P 
and Q coincide at 8; then the straight line becomes the tan- 
gent at 8 ; therefore 

The tangent drawn through the extremity of a diameter is 
parallel to the ordinates of the diameter. 

118. From the focus F draw FC _L to PQ, and let FC 
meet the directrix in the point C. If denote the angle 
which the chord PQ makes with the axis of x, it easily follows 
that DCF— 6 ; then we have 

CP=FP cot0 = — =^t^; that is, 
m A ? ' 

The 'perpendicular to a chord which passes through the focus 
meets the diameter of the chord in the directrix. 

119. Let the tangents drawn through P and Q meet in the 
point T. By solving their equations, 

y<y = 2p(x + xj, 
y 2 y = 2p(x + xj, 

we obtain for the value of the ordinate of T 

2p(x 2 — x l ) 2p yx + 3/2 
y = — — = — o — • Hence 

Tangents drawn through the ends of a chord meet in the 
diameter of the chord. 



138 ANALYTIC GEOMETRY. 

120. What is the locus of the foot of a perpendicular dropped 
from the focus to a tangent f 

Let the equation of the tangent be 

y = mx 4- — • 

* m 

Then the equation of the perpendicular will be 

x p 

* mm 

Since these two lines have the same intercept on the axis 
of y, they meet in that axis ; that is, in the tangent through 
the vertex. This tangent, therefore, is the required locus. 

121. The perpendicular 

x p 
17 mm 

meets the directrix in the point f — p, ). But the ordinate 

P 
of the point where the tangent y — mx -\ meets the para- 

2p m 

bola y 1 == ipx is also — ; therefore 

The perpendicular from the focus to a tangent, if produced, 
meets the directrix in the diameter through the point of contact. 

122. The distance of any point (A, k) from the focus (p } 0) i, f 



and its distance from the point ( — p, — J is 



If (A, Jc) is in the tangent y — mx + 



THE PARABOLA. 139 

P 



then Jc = mh + 

771 

Now this value of Jc, substituted in the two expressions for 
distances just given, makes them equal ; therefore 

Every point in the tangent is equidistant from the focus and 
the point where the perpendicular from the focus to the tangent 
meets the directrix. 

123. What is the locus of the intersection of two perpendicular 
tangents ? 

If the equation of one tangent is 

P 



y — mx + 



m 



then the equation of the other is found to be 

Subtracting one equation from the other, 

(x+p)( K m+-J = 0. 

But mA is not ; therefore 

m 

x+p = 0, 

or x = —p, 

the equation of the directrix. 

Hence the directrix is the locus required. 

124. Tangents are drawn through the ends of a focal chord. 
What is the locus of their intersection f 

Let (h, Jc) be their intersection ; then we may write the 
equation of the chord ly = ^ + A) 



140 



ANALYTIC GEOMETRY. 



If the chord passes through the focus (p, 0), we have 
= 2p(jp + h) ■ 
whence h — — p. 

Therefore the required locus is the directrix. And therefore 
combining this result with that obtained in § 123, we see that 

Tangents through the ends of a focal chord are perpendicular 
to each other, 

125. To find the equation of a parabola referred to any 
diameter and the tangent through its extremity as axes. 

Transform the equation y 2 = Apx to the diameter SX* 
(Fig. 52) and the tangent through 8 as new axes. Let m be 
the slope of the tangent, the angle which the tangent makes 
with the diameter ; then m == tan 0. 




Fig. 52. 



First change the origin to 8 without changing the direction 

of either axis. o 

p Zp 

The co-ordinates of 8 are — v — (§ 116). Therefore the 

m 1 m v ° J 

new equation is 



or 



* m / P V m} 



my 1 -f Apy = Apmx. 



(i) 



THE PARABOLA. 141 

Now retain the axis of x, and turn the axis of y till it coin- 
cides with the tangent at S ; then for any point P we have 

The old x = SP. The new x = SJST. 

The old y = PP. The new y = iVP. 

And it is easily seen from Fig. 52 that 

SP^SN+NPco&e, 

Therefore equation (1) is transferred to the new system by 
writing x + y cos 6 in place of x, and y sin 6 in place of y. 
Making this substitution, and reducing, we obtain 

*=isefv*> - & 

an equation of the same form as y 2 = 4p#. 
Join # to the focus F ; , then 

" ' r l m* m 1 sir 

Therefore equation (2) may be more simply written 

f = 4/*, (3) 

where ^>' is the distance of the origin from the focus. It is 
easy to see that this equation includes the case where the 
axes are the axis of the curve and the tangent at the vertex. 

The quantity 4p' is called the Parameter of the diameter 
passing through S. When the diameter is the axis of the 
curve, it is called the Principal Parameter. 

126. Let the equation of a parabola referred to any diam- 
eter, and the tangent at the end of that diameter as axes, 
be y 2 = 4:p'x. Since the investigations in §§ 109-112 hold 
good whether the axes are at right angles or not, it follows 
immediately that the equation of the tangent at any point 
(#i, Vi) is y<y — 2jo'(# + #i), and that the straight line 

P 1 
y = mx + — will touch the parabola for all values of m. 



142 



ANALYTIC GEOMETRY. 



127. To find the polar equation of a parabola, the focus 
being the pole. 







Fig. 53. 



Let P (Fig. 53) be any point (p, 6) of the curve, and let 6 
be measured from the vertex A of the curve in the same 
direction as clock motion. By definition, 



Now 



Therefore 



or 



FP = PN=DM. 

FP = P, 
DM=DF-MF t 

— 2p — p cos 0. 
p = 2p — p cos 0, 

9 ~ l + cos e* 



[29] 



THE PARABOLA. 143 

Ex. 31. 

1. Prove that the polar of the focus is the directrix. 

2. Prove that the perpendicular dropped from any point 
of the directrix to the polar of the point passes through the 
focus. 

3. To find by construction the pole of a focal chord. 

4. Prove that through any point three normals can be 
drawn to a parabola. 

5. Tangents are drawn through the ends of a chord. 
Prove that the part of the corresponding diameter contained 
between the chord and the intersection of the tangents is 
bisected by the curve. 

6. Focal radii are drawn to two points of a parabola, and 
tangents are then drawn through these points. Prove that 
the angle between the tangents is equal to half the angle 
between the focal lines. 

7. Prove that the locus of the intersection of two tangents 
to the parabola y 2 = ipx, which make an angle of 45°, is the 
parabola y 2 = x 2 + 6px -\-p 2 . 

8. Explain how tangents to a parabola may be drawn from 
an exterior point (§§ 121, 122). 

9. Having given a parabola, how would you find its axis, 
directrix, focus, and latus rectum ? 

10. From the point (— 2, 5) tangents are drawn to the 
parabola y 2 = 6x. What is the equation of the chord of 
contact ? 

11. The general equation of a system of parallel chords in 
the parabola 7y 2 = 25x is 4:X — 7?/ + £ = 0. What is the 
equation of the corresponding diameter ? 



144 ANALYTIC GEOMETRY. 

12. In the parabola y 2 — 13;r, what is the equation of the 
ordinates of the diameter y +11= ? 

13. In the parabola y 2 — 6#, what chord is bisected at the 
point (4, 3) ? 

14. Given the parabola y 2 — Apx ; find the equation of the 
chord which passes through the vertex and is bisected by the 
diameter y ' = a. How can this chord be constructed ? 

15. The latus rectum of a parabola = 16. What is the 
equation of the curve if a diameter at the distance 12 from 
the focus, and the tangent through its extremity, are taken as 
axes? 

16. Show that the equations of that chord of the parabola 
y 2 — ipx which is bisected at the point (h, k) is 

17. Prove that the parameter of any diameter is equal to 
the double ordinate which passes through the focus. 

18. Discuss the form of the parabola from its polar equation. 

19. Show that if the vertex is taken as pole, the polar equa- 
tion of a parabola is 4 # cos $ 

P= sin 2 ' 

20. Find the locus of the foot of a perpendicular dropped 
from the focus to the normal to a parabola. 

21. Two normals to a parabola are perpendicular to each 
other ; find the locus of their intersection. 

22. Find the locus of the centre of a circle which touches a 
given circle and also a given straight line. 

23. The area and base of a triangle being given, find the 
locus of the intersection of perpendiculars dropped from the 
ends of the base to the opposite sides. 



CHAPTER VI. 
THE ELLIPSE. 

Simple Properties of the Ellipse. 

128. The Ellipse is the locus of a point, the sum of whose 
distances from two fixed points is constant. 

The fixed points are called Foci ; and the distance from any 
point of the curve to a focus is called a Focal Eadins. 

The constant sum is denoted by 2 a, and the distance be- 
tween the foci by 2 c. 

c 
The fraction - is called the Eccentricity, and is represented 

by the letter e. Therefore c = ae. 

From the definition of the ellipse it is clear that if 2 a < 2c, 
ora<c, the locus does not exist ; if a = c, the locus is simply 
that part of the straight line joining the foci which is com- 
prised between the foci. The ellipse is the curve obtained 
when a > c ; that is, when e < 1. 

129. To construct an ellipse, having given the foci and the 
constant sum 2 a. 

I. By Motion. Fix pins in the paper at the foci. Tie a 
string to them, making the length of the string exactly equal 
to 2a. Then press a pencil against the string so as to make 
it tense, and move the pencil, keeping the string constantly 
stretched. The point of the pencil will trace the required 
ellipse ; for in every position the sum of the distances from 
the point of the pencil to the foci is equal to the length of the 
string. 



146 



ANALYTIC GEOMETRY. 



II. By Points. Let F, F' be the foci ; then FF' = 2c. 

Bisect FF f at 0, and from lay off OA = OA' = a. 

Then AA'=2a. 

AF +AF ! =(a — c) + 2c + (a-c) = 2a. 
AF' + A'F' - (a - c) + 2c + (a - <?) = 2a. 

Therefore A and J.' are points of the curve. 

Between A and A f mark any point X\ then describe two 
arcs, one with F as centre and AX as radius, the other with 
i* 7 ' as centre and A* X as radius : the intersections F, Q of 




these arcs are points of the curve. By merely interchanging 
the radii, two more points, F, S, may be found. 

After a sufficient number of points has been obtained, draw 
a continuous curve through them. 

130. The line A A' is the Transverse or Major Axis, A, A r 
the Vertices, and the Centre of the curve. 

The line BB\ perpendicular to the major axis at 0, is the 
Conjugate or Minor Axis ; its length is denoted by 2k 

Show that B and B 1 are equidistant from the foci, that 
BF= a, that BO = b, and that a 2 = b 2 + c\ 



THE ELLIPSE. 



147 



131. To find the equation of the ellipse, having given the 
foci and the constant sum 2 a. 




Take the line A A 1 (Fig. 55), passing through the foci, as 
the axis of x, and the point 0, half way between the foci, as 
origin. Let P be any point (x, y) of the curve, and let r, r' 
denote the focal radii of P. Then from the definition of the 
curve, and from the right triangles FPM, P'PIf, 

r's =if+(c + xf 



f + {c-xf 



By addition , r' 2 + r 2 = 2(x 2 + y 2 + c 2 ). 

By subtraction, r' 2 — r 2 = 4 ex. 

Factor (4) , (r \ + r) (r ' — r) = 4 ex. 

2 ex 
Put 2 a for r 1 + r, r\ — r — — — 

Whence 



ex 



Substitute in (3), 



r = a = a — ex. 

a 

, ex 

r = a -\ = a + ex. 

a 



2 (a 2 + c ^) = 2(x 2 + y 2 + c 2 ). 
V a J 

Keduce, and substitute b 2 in place of a 2 — c 2 (§ 130), 



or 



« 2 , y* _ - 

^2 + p- 1 ' 



(1) 

(2) 
(3) 
(4) 
(5) 

(6) 
(7) 
(8) 

(9) 



[30] 



148 ANALYTIC GEOMETRY. 

132, To trace the form of the curve from its equation. 

The intercepts on the axis of x are + a and — a ; on the 
axis of y, -\-b and — b. 

Only the squares of the variables x and y appear in the 
equation ; hence, if it is satisfied by a point (#, y), it will also 
be satisfied by the points (x, —y), (— #, y), (—#, — y). There- 
fore we infer that 

(i.) The curve is symmetrical with respect to the axis of x. 

(ii.) The curve is symmetrical with respect to the axis of y. 

(iii.) Every chord which passes through the point is bisected 
at ; for the distance from either (x, y) or (—x, — y) to (0, 0) 
is V# 2 + 2/ 2 . This explains why is called the centre. 

Since the sum of f - J and f V J is 1, neither of these squares 

can exceed 1 ; therefore the maximum value of x is -{-a, and 
the minimum value — a, while the corresponding values of y 
are + b and — b. Therefore the curve is wholly contained 
within the rectangle whose sides are equal to 2a and 2b, and 
are bisected by the axes. 

133. To trace the changes in the form of the ellipse when the 
semi-axes are supposed to change. 

Let a be regarded as a constant, and b as a variable. 

(i.) Suppose b to increase. Then c decreases (since c 2 = a 2 
— b 2 ), e decreases, the foci approach the centre, and the ellipse 
approaches the circle. 

(ii.) Let b — a. Then c = 0, e = 0, the foci coincide with 
the centre, the ellipse becomes a circle of radius a, and equa- 
tion [30] becomes , , , 

L J x 2 + y 2 = a 2 . 

Therefore we may regard a circle as an ellipse whose eccen- 
tricity is equal to 0. 



THE ELLIPSE. 149 

(iii.) Let b > a. The foci and major axis will now be on the 
axis of ' y, c will increase with b, e= -> and b 2 = a 2 + c 2 . 

(iy.) If we suppose 6 to decrease to (a remaining con- 
stant), c will increase to a, e will increase to 1, while the 
curve will approach, and finally coincide with, the major axis, 
its equation at the same time becoming y = 0. 

134. It follows from § 131 that a point (h, k) is on the 
ellipse represented by equation [30], provided 

h 2 P 

- + - - 1 = 0. 
a 2 b 2 

It may be shown by reasoning similar to that employed in 

§ 105 that the point (h, Jc) is outside or inside the curve, 

h 2 lc 2 
according as — + j: 2 ~ 1 is positive or negative. 

135. Since the constants a and b in the equation 

- + ^ = 1 (1) 
a 2 ^6 2 V ; 

may have any positive values, every equation reducible to the 

form of (1) must represent an ellipse. Hence every equation 

of the form A*+B£ = C 

represents an ellipse, provided Cis not zero, and A, B, and 
all have the same sign. Its semi-axes have the values 

a= \'Z b= \B 

136. The chord passing through either focus perpendicular 
to the major axis is called the Latns Kectnm or Parameter. 

To find its length, put x = c in the equation of. the ellipse. 

Then tf= b (a 2 -c 2 )=% y = ±-- ' 

a 2 a 2 a 

2 b 2 

Therefore the latus rectum = 

a 



150 



ANALYTIC GEOMETRY. 



137. The circle having for diameter the major axis of the 
ellipse is called the Auxiliary Circle ; its equation is 

a? + y 2 = a\ 

The circle having for diameter the minor axis is called the 
Minor Auxiliary Circle ; its equation is 

x 2 + y 2 = b 2 . 

If P (Fig. 56) is any point of an ellipse, and the ordinate 
MP produced meets the auxiliary circle in Q, the point Q is 
said to correspond to the point P. 

The angle QOMis called the Eccentric Angle of the point P, 
and denoted by the letter <£. 




138. Let y, y f represent the ordinates of points in an ellipse 
and the auxiliary circle respectively, corresponding to the 
same abscissa x. Then from the equations of the two curves 
we have -, 



y = - Va 2 — x 2 , y f = Va 2 — x 2 . 



Whence 



y-y 



= b 



a, or 



The ordinates of the ellipse and the auxiliary circle, corre- 
sponding to a common abscissa, are to each other in the constant 
ratio of the semi-axes of the ellipse. 



THE ELLIPSE. 151 

139, Hence, if the axes 2a, 2b of an ellipse are given, we 
may find any number of points in the ellipse by constructing 
the auxiliary circle, drawing ordinates at pleasure, and then 
reducing their lengths in the ratio b : a. 

In practice, it is convenient to proceed as follows : 
Construct both the major and minor auxiliary circles ; draw 
any radius, cutting the circles in Q, E, respectively ; through 
Q draw a line II to Y, and through E draw a line II to OX: 
the intersection P of these parallels is a point of the ellipse. 
For from the similar triangles QOM, OEN, 

ON: QM = OE:OQ. 

Now ON^ PM= y, QM= y f , 

OE = b, OQ =a. 

Therefore y : y f = b : a. 

With the aid of the eccentric angle <f> = QOX, the proof 
that P is a point of the ellipse may be given as follows : 
Let P be the point (x, y) ; then 

x = OM— OQ cos<f>~a cos <j>, 

y = PM= 0N= OE sin <f> = b sin <£. 

Whence we have 



Square 

Add 

But 

Hence 

Therefore P is a point of the required ellipse. 



X 

a 


= COS ' 


4>, 


y 
b~ 


sin <£. 




X 2 

a 2 


= COS 1 


'■^ 


f 
b 2 


= sin 2 


0. 


X 2 

a* 


^b 2 


= cos 2 <£ + sin 2 


4> 


cos 2 <f> + 


sin 


'<£= 


= 1. 




3? 

a 2 


^ b 2 


= 1, 









152 



ANALYTIC GEOMETRY. 



140. Another mode of constructing an ellipse from its axes 
is shown in Fig. 57. 

In the rectangle OACB, whose sides OA, OB are made 
equal to the given semi-axes a and 5, divide the side BC 
into any number of equal parts, and divide BO into the same 
number of equal parts, and let M, JY denote any two corre-, 
sponding points of division, counting from B. If we now draw 
through the extremities A, A' of the major axis, and the points 
M t iV, respectively, straight lines, the intersection P of the 
lines will be a point of the required ellipse. 




In order to give a general proof, let there be n equal parts, 
and let ON and CM contain r of these parts, respectively ; 

then rb ra 

om=-, CM=-- 

n n 

Produce MN to meet OB produced in Q; then 
OQ:AC=OA:CM } 
ra 



OQ: b= a: 



n\r. 



Therefore 



OQ 



nb 



x -+ 

a 


ry_ 
nb 


= 1, 


x -+ 

a 


ny 
~rb 


= 1. 


nb 


= 1 


a; 

> 

a 


7%/ = 

rb 


a 


b 2 


= 1- 


a; 2 
a 2 


a 2 ^ 




= 1. 



THE ELLIPSE. 153 

Taking now for origin, and OA for axis of x, we have for 
the symmetrical equations of AM and J.'iV 



and 

Or 
and 

Multiply 

that is, 

This relation must hold true of the point common to the 
lines AM and A'N) therefore this point is on the ellipse 
whose axes are 2a and 2b. 

Ex. 32. 

What are a, b, c, and e in the ellipse w T hose equation is 

1 £!-lJ£ = 1? 

' 25 16 

2. x 2 + 2y 2 = 2? 

3. 3^ 2 + 4y 2 = 12? 

4. ^4^ 2 +^ 2 -l? 

5. Find the latus rectum of the ellipse 3# 2 + 7y 2 =18. 

6. Find the eccentricity of an ellipse if its latus rectum is 
equal to one-half its minor axis. 



154 ANALYTIC GEOMETRY. 

What is the equation of an ellipse if 

7. The axes are 12 and 8 ? 

8. Major axis = 26, distance between foci = 24 ? 

9. Sum of axes = 54, distance between foci = 18 ? 

10. Latus rectum = - 6 ^, eccentricity = f ? 

11. Minor axis = 10, distance from focus to vertex = 1 ? 

12. The curve passes through (1, 4) and (—6, 1) ? 

13. Major axis = 20, minor axis = distance between foci? 

14. Sum of the focal radii of a point in the curve = 3 times 
the distance between the foci ? 

15. Prove that the semi-minor axis is a mean proportional 
between the segments of the major axis made by one of the 
foci. 

16. What is the ratio of the two axes if the centre and foci 
divide the major axis into four equal parts ? 

17. For what point of an ellipse is the abscissa equal to the 
ordinate ? 

Find the intersections of the loci 

18. 3* 2 + 6y 2 = ll and y = x + \. 

19. 2£ 2 + 3y 2 = 14 and y 2 = ±x. 

20. x 2 + ly 2 = l§ and x 2 + y 2 = 10. 

21. The ordinates of the circle x 2 -\-y 2 =r 2 are bisected; 
find the locus of the points of bisection. 

22. A straight line AB so moves that the points A and B 
always touch two fixed perpendicular straight lines. Show 
that any point P in AB describes an ellipse, and find its 
equation. 






THE ELLIPSE. 155 

23. What is the locus of Ax 2 +By 2 = when C is zero ? 
When is this locus imaginary ? 

24. Prove that the abscissas of the ellipse b 2 x 2 + a 2 ?/ 2 = a 2 b 2 
are to the corresponding abscissas of the minor auxiliary circle, 

# 2 + y 2 — ^ 2 > as a : £• 

25. Construct an ellipse by the method of § 139. 

26. Construct an ellipse by the method of § 140. 

27. Construct the axes of an ellipse, having given the foci 
and one point of the curve. 

28. Construct the minor axis and foci, having given the 
major axis (in magnitude and position) and one point of the 
curve. 

29. A square is inscribed in the ellipse 

Find the equations of the sides and the area of the square. 

Tangents and Normals. 

141. To find the equations of a tangent and a normal to an 
ellipse, having given the point of contact (x x , y x ). 

Taking the equation of the ellipse, 

b V + a\f = a 2 b\ 

and the equation of the straight line through (x^ y x ) and 

O;, 2/ 2 ), y — y x y 2 — y x 



■v JL\ X>> X' 



and proceeding as in § 72, we obtain as the equation of a 
chord through (x^ y x ) and (x 2 , y 2 ) 

y — Vi = h 2 (x 1 + x 2 ) 
x — x x a\y x + y 2 ) 



156 ANALYTIC GEOMETRY. 

Now make x 2 = x u 3/2 = 3/1; then the chord becomes a tan- 

gen ' an y~Vi __ _ 6 2 Qi + z 2 ) 

x-x x a\y x + y) 

-u y — V\ b 2 x x 

becomes £ — — S 

x — x x d L y x 

which reduces to 5? + M = 1. [31] 

a 2 ft 2 L J 

From the equation above it appears that the value of the 
slope of the tangent, in terms of the co-ordinates of the point 
of contact is 7 2 

O X\ 

The normal is perpendicular to the tangent, and passes 
through (xi, y x ) ; therefore its equation is easily found (by the 
method of § 51) to be 

y x oo oc x y ocsj^a 2 — b 2 ) 



b 2 a 2 a 2 b 2 



[32] 



142. To find the sub tangent and subnormal. 

Making y = in [31] and [32], and then solving the equa- 
tions for x, we obtain : 

a 2 
Intercept of tangent on axis of x = — > 



Intercept of normal on axis of x = — x x = e 2 x x . 

a 2 

Whence the values of the subtangent and the subnormal 
(denned as in § 71) are easily found to be as follows : 

Subtangent = — — — -9 [33] 

b 2 
Subnormal = — ac x . [34] 



THE ELLIPSE. 



157 



143. If tangents to ellipses having a common major axis are 
drawn at points having a common abscissa, they will meet on 
the axis of x. 

For in all these ellipses the values of a and x are constant, 
and therefore (by § 142) the tangents all cut the same inter- 
cept from the axis of x. 




■9) 



^ 



V 



Fig. 58. 



\ 3 



144. The normal at any point of an ellipse bisects the angle 
formed by the focal radii. 

The values of the focal radii for the point P (Fig. 58) were 
found in § 131 to be 

PF= a - ex lt PF f = a + ex x . 
If the normal through P meets the axis of x in N, ON=e 9 x 
(§ 142) ; and therefore 

NF = c — e 1 x l = ae — e 2 x x — e(a — ex), 
]\ r F' = c -f- e 2 x x = ae + e 2 x 1 = e(a + ex). 

Therefore NF : NF' ' = PF ': PF\ 

or the normal divides the side FF f of the A PFF' into two 
parts proportional to the other two sides. Therefore (by- 
Geometry) FPN=F'PN. 

The tangent FT, being perpendicular to the normal, must 
bisect the angle FPR, formed by one focal radius with the 
other produced. 



* 



158 



ANALYTIC GEOMETRY. 



145. To draw a tangent and a normal through a given point 
of an ellipse. 

I. Let P (Fig. 59) be the given point. Describe the 
auxiliary circle, draw the ordinate PM, produce it to meet 
the circle in Q, draw QT tangent to the circle and meeting 
the axis of x in T, and join PT] then PT is a tangent 
to the ellipse (§ 143). Draw PN ± to PT] PN is the 
normal at P. 

C 




II. Draw the focal radii, and bisect the angles between 
them. The bisectors are the tangent and the normal at the 
point P (§ 144). 

146. To find the equation of a tangent to an ellipse, having 
given its direction. 

This problem may be solved by finding under what condi- 
tion the straight line , ,-,, 
b y = mx + c (1) 

will touch the ellipse b 2 x 2 + a 2 \f — a 2 b 2 . (2) 

Eliminating y from (1) and (2), and then solving for x } we 
find two values of x : 

— m a 2 c zt ah-yjvra 1 -\-b 2 — c 2 



lira 1 + b 2 
These values will be equal if 



m 2 a 2 + b 2 — c 2 = 0, or c = =b VmV -f- b 2 . 



THE ELLIPSE. 159 

If the two values of x are equal, the two values of y must' 
also be equal from equation (1). 

Therefore the two points in which the ellipse is cut by the 
line will coincide if c = zb VmW + b 2 . 

Hence the straight line 

y = moc ± (m?a* + 6 2 ) [35] 

will touch the ellipse for all values of m. 

Since either sign may be given to the radical, it follows that 
two tangents may be drawn to an ellipse in a given direction 
(determined by the value of ra). 

147. To find the locus of the intersection of two tangents to 
an ellipse which are perpendicular to each other. 

Let the equations of the tangents be 



y — nix + Vw 2 a 2 + b 2 , (1) 

y = m'x + Vra'a 2 -f- b 2 . (2) 

The condition to be satisfied is 

i - i 

mnv — — 1, or m ■= 

m 

If we substitute for ?n f in equation (2) its value in terms of 

m, the equations of the tangents may be written 

y — mx — Vm 2 a 2 + b 2 , (3) 

my -h x = Va 2 + ?n 2 b 2 . (4) 

The co-ordinates, x and y, of the intersection of the tangents 

satisfy both (3) and (4) ; but before we can find the constant 

relation between them we must first eliminate the variable m. 

This is most easily done by adding the squares of the two 

equations ; the result is 

(1 + m 2 )x 2 + (1 + m 2 )y 2 — (1 + m 2 ) (a 2 + 5 2 ), 
or x 2 -f y 2 — a 2 + b 2 . 

The required locus is therefore a circle. This circle is 
called the Director Circle of the ellipse. 



160 ANALYTIC GEOMETRY. 



Ex. 33. 



1. What are the equations of the tangent and the normal 
to the ellipse 2x 2 -{-3y 2 = 35 at the points whose abscissa = 2? 

2. What are the equations of the tangent and the normal to 
the ellipse 4# 2 + 9?/ 2 = 36 at the points whose abscissa = — -f ? 

3. Find the equations of the tangent and the normal to the 
ellipse x 2 + 4y 2 = 20 at the point of contact (2, 2). Also find 
the subtangent and the subnormal. 

4. Show that the line y = x + Vf touches the ellipse 

5. Required the condition which must be satisfied in order 
that the straight line ■ \- — = l may touch the ellipse 

a 2 ^b 2 

6. In an ellipse the subtangent for the point (3, - 1 -^ 2 -) is 16, 
the eccentricity = 4. What is the equation of the ellipse ? 

7. What is the equation of a tangent to the ellipse 
9^ 2 + 64y 2 =--576 parallel to the line 2y = x? 

8. Find the equation of a tangent to the ellipse Sx 2 -\-5y 2 — 15 
parallel to the line 4# — 3y — 1 = 0. 

9. In what points do the tangents which are equally inclined 
to the axes touch the ellipse b V + a 2 y 2 = a 2 b 2 ? 

10. Through what point of the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 
must a tangent and a normal be drawn in order that they 
may form, with the axis of x as base, an isosceles triangle ? 

11. Through a point of the ellipse b 2 a? + a 2 y 2 = a 2 b 2 , and 
the corresponding point of the auxiliary circle x 2 + y 2 = a 2 , 
normals are drawn. What is the ratio of the subnormals ? 



THE ELLIPSE. 161 

12. For what points of the ellipse b 2 x 2 + ahj 2 = a 2 b 2 is the 
subtangent equal numerically to the abscissa of the point of 
contact ? 

13. Find the equations of tangents drawn from the point 
(3, 4) to the ellipse 16 x 2 + 2hy 2 = 400. 

14. What are the equations of the tangents drawn through 
the extremities of the latera recta of the ellipse 4# 2 +9?/ 2 =36a 2 ? 

15. What is the distance from the centre of an ellipse to a 
tangent making the angle <£ with the major axis ? 

16. What is the area of the triangle formed by the tangent 
in the last parabola and the axes of co-ordinates ? 

17. From the point where the auxiliary circle cuts the 
minor axis produced tangents are drawn to the ellipse. Find 
the points of contact. 

18. Prove that the tangents drawn through the ends of a 
diameter are parallel. 

19. Find the locus of the foot of a perpendicular dropped 
from the focus to a tangent. 

Ex. 34. (Review.) 

1. Given the ellipse 36.r 2 + IOO3/ 2 = 3600. Find the equa- 
tions and the lengths of focal radii drawn to the point (8, ^-). 

2. Is the point (2, 1) within or without the ellipse 
2.r 2 + 3y 2 --12? 

Find the eccentricity of an ellipse 

3. If the equation is 2x 2 -f- 3y 2 = 1. 

4. If the angle FBF 1 = 90° (see Fig. 54). 

5. If LFR is the latus rectum and LOR is an equilateral 
triangle {F being the focus, O the centre). 



162 ANALYTIC GEOMETRY. 

Find the equations of tangents to an ellipse 

6. If they make equal intercepts on the axes. 

7. If they are parallel to BF (Fig. 54). 

x 1/ 

8. "Which are parallel to the line - + t = 1 (a and b being 
the semi-axes). 

9. Find the equation of a tangent in terms of the eccentric 
angle <f> of the point of contact. 

Find the distance from the centre of an ellipse to 

10. A tangent through the point of contact (x Yl y{). 

11. A tangent making the angle <j> with the axis of x. 

12. In what ratio is the abscissa of a point divided by the 
normal at that point ? 

13. At the point (x^ y x ) of an ellipse a normal is drawn. 
What is the product of the segments into which it divides the 
major axis? 

14. Find the length of P If (Fig. 58). 

15. Determine the value of the eccentric angle at the end 
of the latus rectum. 

Prove that the semi-minor axis b of an ellipse is a mean 
proportional between 

16. The distances from the foci to a tangent. 

17. A normal and the distance from the centre to the cor- 
responding tangent. 

Determine and describe the loci of the following points : 

18. The middle point of that portion of a tangent contained 
between the tangents drawn through the vertices. 

19. The middle point of a perpendicular dropped from a 
point of the circle (x — a) 2 + y 2 = r 2 to the axis of y t 



THE ELLIPSE. 163 

20. The middle point of a chord of the ellipse b 2 x 2 -\-a 2 y 2 —a 2 b 2 
drawn through the positive end of the minor axis. 

21. The vertex of a triangle whose base, 2 c, and sum of 
the other sides, 2s, are given. 

22. The vertex of a triangle, having given the base 2 c and 
the product k of the tangents of the angles at the base. 

23. The symmetrical point of the right-hand focus of an 
ellipse with respect to a tangent. 



SUPPLEMENTARY PROPOSITIONS. 

148. Two tangents can be drawn to an ellipse from any 
point ; and they will be real, coincident, or imaginary, accord- 
ing as the point is outside, on, or inside the curve. 

If the tangent y — mx -f- "Vm 2 a 2 + b 2 pass through the point 
(A,*), then * = mA + VmV + * f , 

or (A 2 - a 2 )m 2 — 2 hkm + k 2 -b 2 = 0. 

This is a quadratic equation with respect to m, and its roots 
give the directions of those tangents which pass through (h, k). 
Since a quadratic equation has two roots, two tangents may 
be drawn from any point (h, Tc) to an ellipse. 

By solving the equation, we obtain 



Kk ± Vb 2 /i 2 + a 2 k 2 - a 2 b 2 . 
A 2 — a 2 

and we see that the roots are real, coincident, or imaginary, 

according as b 2 h 2 + a 2 k 2 — a 2 b 2 is positive, zero, or negative ; 

h 2 F 
that is, according as — + 7:, — 1 is positive, zero, or negative ; 

in other words, according as the point (h, 1c) is outside, on, 
or inside the ellipse. (§ 134.) 



164 



ANALYTIC GEOMETRY. 



149. To find the equation of the straight line passing through 
the points of contact of the tivo tangents drawn to an ellipse from 
the point (A, 1c). 

If (#!, yx) and (x 2 , y 2 ) are the points of contact, it follows, 
by reasoning similar to that employed in § 78, that the 

required equation is 

hx , £y_-, 
a 2 " 1 " b 2 ~ 1 ' 

This line is always real ; but if the point (h, 1c) is within 
the ellipse, the points (x x , i/i), (x 2 , y 2 ) through which the line 
passes, will be imaginary. 

150. The straight line joining the points of contact of the 
two tangents from any point P to an ellipse is called the Polar 
of P with respect to the ellipse ; and the point P is called 
the Pole of the straight line with respect to the ellipse. 







Fig. 60. 



The propositions in §§ 80-82, relating to poles and polars 
with respect to a circle, also hold true for poles and polars 
with respect to an ellipse, and may be proved in the same 
way. 



THE ELLIPSE. 165 

151. To draw a tangent to an ellipse from a given point P 
outside the curve. 

Suppose the problem solved, and let the tangent meet the 
ellipse at Q (Fig. 60). If F'Q be produced to G, making 
QG = QF, then A FQG is isosceles ; now Z FQF^Z GQF 
(§ 144) ; therefore FQ is perpendicular to FG at its middle 
point ; therefore F is equidistant from F and G. This re- 
duces the problem to determining the point G. 

Since F'G = 2a, G lies in the circle with F* as centre and 
2a as radius. And G also lies in the circle with F as centre 
and FF as radius. Hence the construction is obvious. 

152. To find the locus of the middle points of the system of 
chords represented by the equation 

y = mx + k. (1) 

Let any one of the chords meet the ellipse 

b V + a 2 y 2 = a 2 b 2 (2) 

in the points (x x , y^) and (x 2 , y 2 ) ; then (§ 141 (3)) 

b\x x + x 2 ) 



d\y! + y 2 ) 
If (x,y) is the middle point, 2x = x x -{- x 2l 2y = y l -\-y 2} 



m = -> (4) 

a 2 y 

b 2 x /CN 

or y = - — . (5) 

a z m 

This relation holds true for the middle points of all the 
chords ; therefore it is the equation required. 

The locus of the middle points of a system of parallel chords 
in an ellipse is called a Diameter of the ellipse. 

From the form of (5) we see that a diameter is a straight 
line passing through the centre. 



166 ANALYTIC GEOMETRY. 

153. Let m! denote the slope of the diameter of the chords 
represented by the equation y — mx-\-k ; then from (5) of § 151 

mm' = — ^. [36] 

From the symmetry of this equation we may infer at once 
that all chords parallel to the diameter y — m'x are bisected 
by the diameter y = mx ; hence 

If one diameter bisect all chords parallel to another, the 
second diameter bisects all chords parallel to the first. 

Two such diameters are called Conjugate Diameters, 

154. Let a straight line cutting the case in P and Q move 
parallel to itself till P and Q coincide with the end of the 
diameter bisecting PQ ; then the straight line becomes the 
tangent at the end of the diameter. Therefore 

A diameter bisects all chords parallel to the tangent at its 
extremity. A tangent drawn through the end of a diameter 
is parallel to the conjugate diameter (§ 153). 

155. Let POP, POP' (Fig. 61) be two conjugate diameters, 
meeting the curve on the positive side of the axis of x in the 
points P (x x , 2/x) and P (x 2 , y 2 ), and making the angles a, /2 
respectively with the axis of x. 

Let a be acute ; then it follows from equation [36] that /? 
must be obtuse ; whence we infer that two conjugate diameters 
must lie in different quadrants. 

The equation of the tangent through P is 

a 1 ^ V K J 

Therefore the equation of the diameter POP 1 , which is 
parallel to this tangent (§ 154) and passes through 0, is 

a2 + b 2 - 0. W 



THE ELLIPSE. 



167 



Similarly, the equation of POP 1 is found to be 

-0. 



X 2% j I 



The point R is in the locus of (2) ; therefore 

= 0. 



X\X>i 

IF 



Mi 



(3) 



(4) 



Equation (4) is the condition which must be satisfied by 
the co-ordinates of the extremities of every pair of conjugate 
diameters. 




Fig. 61. 

156. Let the ordinates of the extremities P, P (Fig. 61) of 
two conjugate diameters meet the auxiliary circle in Q, S 
respectively, join QO and SO, and denote Z QOX by <£, 
Z SOX by <£'. Then the values of the co-ordinates of P 
and R are (§ 138) 

x Y = a cos <£, x 2 = a cos <£', 

yi = b sin <£, y 2 = b sin </>'. 

Whence, by substitution in equation (4) of § 155, we obtain 

cos <£ cos <f> f + sin <£ sin <f>' = 0. 
Therefore <f>' -</>= iir. 

That is, the difference of the eccentric angles corresponding 
to the ends of two conjugate diameters is equal to a right angle. 



168 ANALYTIC GEOMETRY. 

157. Given the end (x u y x ) of a diameter, to find the end 
(x 2 , Vi) of the conjugate diameter. 

From § 156 we have for one of the ends 

x 2 = a cos <£' — a cos (<£ -f £ tt) = a sin <£, 

y 2 = b sin <£' — b sin (<£ + \-n) = b cos <£. 

x 2 — a sin <£ a y 2 b cos </> J 

2/i ~~ b sin </> 6' x 1 a cos </> a* 

Therefore x 2 = — -y^, y 2 = - a^. 

Since every chord through the centre is bisected by the 
centre, the co-ordinates of the other end of the diameter are 

a , b 

-y x and x x . 

b a 

158. To find the angle formed by two conjugate semi-diam- 
eters, whose lengths a f , b f are given. 

Let the semi-diameters make the angles a, /? respectively 
with the axis of x, and let denote the required angle. Then 
if (x h y{) and (x 2 , y 2 ) are the extremities of a' and b* respect- 



ively, 



sm« = -„ Bmfi = v = w . 

Xi x 2 ayi 

COS a = -,, COS p — — Ti — ~~ TTT 

a' ' o' bo 

sin = sin (/3 — a) 

= sin /3 cos a — cos /? sin a 
a 2 6 2 



aba f b' 
ab 



THE ELLIPSE. 169 

159. The lines joining any point of an ellipse to the ends 
of any diameter are called Supplemental Chords. 

Let PQ, P'Q be two supplemental chords (Fig. 62). Through 
the centre draw OP parallel to P'Q, and meeting PQ in P ; 
also OP' parallel to PQ, and meeting PQ in P' . 




Fig. 62. 

Since is the middle point of PP \ and OR is drawn 
parallel to PQ, and OP 1 is drawn parallel to P'Q, P and 
P' are the middle points of QP, QP' respectively. Therefore 
OP will bisect all chords parallel to QP, and OP' will bisect 
all chords parallel to QP 1 . Hence OP, OP' are conjugate 
diameters. 

Therefore the diam,eters parallel to a pair of supplemental 
chords are conjugate diameters. 

160. To find the equation of an ellipse referred to a pair of 
conjugate diameters as axes. 

The origin not being altered, we must substitute for x and y 
expressions of the form ax -)- by, a'x + b'y (§ 95). 

Therefore the transformed equation will have the form 



170 ANALYTIC GEOMETEY. 

But by hypothesis the axis of x bisects all chords parallel 
to the axis of y ; therefore the two values of y corresponding 
to each value of x will be equal and of opposite signs. 

Hence C= 0, and the equation 

Ax 2 + Cxy + £y 2 = l 

becomes Ax 2 + By 2 = 1. 

The intercepts of the curve on the new axes are equal to 
the semi- conjugate diameters. If we denote them by a' and 
V, we have iy iy 

whence A = — * £ = — > 

a n b n 

and the equation ^ 2 , g 2 __ ^ 

becomes ^^V*~ t^l 

This equation has the same form as the equation referred 
to the axes of the curve ; whence it follows that formulas 
derived from equation [30], by processes which do not pre- 
suppose the axes of co-ordinates to be rectangular, hold true 
when we employ as axes two conjugate diameters. 

For example, the equation of a tangent at the point (x x , y x ), 
referred to the semi-conjugate diameters a f and b', is 

vw^yjl — ^ 

a n + b' 2 

161. To find the conditions under which an equation will 
represent an ellipse ivhen of the form 

Ax 2 + ~By 2 +Dx + Ey + F= 0. 
If neither A nor B is zero, we may write the equation 



THE ELLIPSE. 171 

If we take a new origin with parallel axes at the point 

( » V and denote by K the constant quantity 

\2A 2 BJ 

which forms the right side of the last equation, the equa- 
tion becomes Ax*+Btf=K, 

or K X y ' 

X 2 V 2 

that is, — -f — = 1, 

A if 

which we know (§ 135) represents an ellipse, provided K be 
not zero and A, B, and iThave like signs. 

K K 

If the denominators — and — are both negative, it is clear 
A B J 

that no real values of x and y will satisfy the equation : the 
locus in this case is called an imaginary ellipse. 

K TC 

It is obvious that the two denominators — and — will have 

A B 

like signs when A and B have like signs ; and by comparing 
the signs of the constants which enter into the value of K, it 
appears that the common sign of the denominators will be 
positive or negative, according as the sign of F is unlike or 
like that of A and B. Hence an equation of the form 

Ax 2 + By 2 + Dx+Ey + F= 

(i.) will represent an ellipse, if A and B are neither of them 
zero and agree in sign. 

(ii.) The ellipse will be real or imaginary, according as the 
sign of F is unlike or like that of A and B. 

The axes of the ellipse are parallel to the axes of co-ordi- 

, 

2A 2B, 

The major axis will be parallel to the axis of x or to the 

axis of y, according as A is less than, or greater than, B. 



172 



ANALYTIC GEOMETRY. 



162. To find the locus of a point which moves so that the 
ratio of its distances from a fixed point and a fixed straight line 
is constant and less than unity. 




Let e denote the constant ratio, 2p the distance from the 
fixed point i^(Fig. 63) to the fixed line CE. Taking CE for 
the axis of y, and the perpendicular to CE through F for the 
axis of x, then from the definition of the locus 



FP=exNP-- 

Therefore we have the relation 



: ex. 



(x-2p) 2 + y 2 =e 2 x 2 , 



or 



(1 - e 2 )x 2 + y 2 - ±px + 4:p 2 = 0. (1) 

Since we suppose <?<1, the coefficients of x 2 and y 2 are 
both positive ; therefore the locus is an ellipse (§ 161). Com- 
paring the coefficients of (1) with those of the first equation 
of (§ 161), we obtain 

A = l-e>, £ = 1, C=~4p, Z> = 0, E=±p 2 . 

Therefore the centre of the locus is the point ( — S- 

\1 — i 







THE ELLIPSE. 173 

Changing the origin to the centre, we obtain an equation 
which may be written in the form 

V = l- (2) 



2ep J (2ep) 

By putting x and y successively equal to 0, we find for the 

values of the semi-axes 

2 ep 7 2ep 

1-e 2 Vl - e 2 

Whence, by substitution in (2), we get the equation in the 
ordinary form 2 2 

Since OD = " t and FD — 2p, therefore 
1 - e 2 l 

1 — er 1 1 — e 2 
But the distance from the centre to the focus of the 

elli P seis . Oe 2 v 

1— e 2 

Therefore the fixed point F coincides with the focus of the 
ellipse. 

Also — = e, or the constant ratio e is equal to the eccen- 
a 

tricity of the ellipse as defined in § 128. 
Whence an ellipse is often defined as 

The locus of a point which moves so that the ratio of its dis- 
tances from a fixed point and a fixed straight line is constant 
and less than unity. 

F\s called the locus ; DN, the Directrix. 

The symmetry of the curve with respect to the minor axis 
shows that there is another focus and another directrix on the 
other side of the minor axis, at distances from it equal respec- 
tively to those of F and CE. 



174 



ANALYTIC GEOMETRY. 



163. To find the polar equation of the ellipse, the right-hand 
focus being taken as the pole. 




Fig. 64. 

Let P be any point (p, 6) of the curve. Then in the 
triangle FFF f 

PF f2 =PF 2 + FF' 2 + 2PFxFF' X cos 6. 

But FF= p, FF 1 = 2c, and by the definition of the curve 
PF' = 2a-p; therefore 

(2 a - p) 2 = p 2 + 4 c 2 + 4 cp cos 0. 

Reducing, and substituting a 2 e 2 for c 2 , we obtain 

a 2 — c 2 a(l — e 2 ) 



a + c cos 6 1 + e cos 9 



[38] 



We may obtain this result more simply by using for the 
focal radius FF the value (§ 131) 

p = a — ex. 
But the general value of x (6 being acute) is 
x = c + p cos 6 = ae + p cos 0. 

By substituting this value of a? we obtain the same polar 
equation as before. 



THE ELLIPSE. 



175 



164. To find the area of an ellipse. 

Divide the semi-major axis OA (Fig. 65) into any number 
of equal parts, through any two adjacent points of division 
M, N erect ordinates, and let the ordinate through M meet 




the ellipse in JP and the auxiliary circle in Q. Through P, Q 
draw parallels to the axis of x, meeting the other ordinate in 
P, S, respectively. Then (§ 138) 

area of rectangle MPRN ^ MP = S # 
area of rectangle MQSN MQ a 

And a similar proportion holds true for every correspond- 
ing pair of rectangles. 

Therefore, by the Theory of Proportions, 

sum of rectangles in ellipse _ b 
sum of rectangles in circle a 

This relation holds true however great the number of rect- 
angles. The greater their number, the nearer does the sum of 
their areas approach the area of the elliptic quadrant in one 
case, and the circular quadrant in the other. In other words, 
these two quadrants are the limits of the sums of the two 



176 ANALYTIC GEOMETRY. 

series of rectangles. Therefore, by the fundamental theorem 

area of elliptic quadrant _ b 
area of circular quadrant a 

Multiplying both terms of the first ratio by 4, 

area of the ellipse b 

area of the circle a 

But the area of the circle = -n-a 2 ; therefore 

area of the ellipse = nab. [39] 

Ex. 35. 

1. What are the equations of the directrices (§ 162) ? 

2. Prove that the polars of the foci are the directrices. 

3. What is the equation of the polar of the point (5, 7) 
with respect to the ellipse 4^ 2 + 9y 2 = 36 ? 

4. Prove that a focal chord is perpendicular to the line 
which joins its pole to the focus. In what line does the pole 
lie? 

5. Find the pole of the line Ax + By + C= with respect 
to the ellipse b 2 x 2 + a 2 y 2 = d 2 b 2 . 

6. Each of the two tangents which can be drawn to an 
ellipse from any point on its directrix subtends a right angle 
at the focus. 

7. The two tangents which can be drawn to an ellipse from 
any external point subtend equal angles at the focus. 

8. Find the slope m of a diameter if the square of the 
diameter is (i.) an arithmetic, (ii.) a geometric, (iii.) an har- 
monic mean between the squares of the axes. 

9. Given the length 21 of a diameter, its inclination 6 to 
the axes, and the eccentricity ; find the major and minor axes. 



THE ELLIPSE. 177 

10. Tangents are drawn from (3, 2) to the ellipse rr 2 +4?/ 2 ^4. 
Find the equation of the chord of contact, and of the line 
which joins (3, 2) to the middle point of the chord. 

11. Find the equation of a diameter parallel to the normal 
at the point (x u ?/ x ), the semi-axes being a and b. 

12. Find the area of the rectangle whose sides are the two 
segments into which a focal chord is divided by the focus. 

13. What is the equation of a chord in the ellipse 
13 .r 2 -f- lly 2 — 143 which passes through (1, 2) and is bisected 
by the diameter 3x — 2y = ? 

14. In the ellipse 9# 2 + 36y 2 = 324 find the equation of a 
chord passing through (4, 2) and bisected at this point. 

15. Write the equations of diameters conjugate to the fol- 
lowing lines : 

x ~ V ~ 0, x -j- y — 0, ax — by, ay = bx. 

16. Show that the lines 2x — y — 0, x + oy = are con- 
jugate diameters in the ellipse 2# 2 + 3y 2 — 4. 

17. If a', V are two semi-conjugate diameters, prove that 
a* + j» = a » + b\ 

18. The area of the parallelogram formed by tangents 
drawn through the ends of conjugate diameters is constant, 
and equal to 4a5. 

19. The diagonals of the parallelogram in No. 18 are also 
conjugate diameters. 

20. The angle between two semi-conjugate diameters is a 
minimum, when they are equal. 

21. The eccentric angles corresponding to equal semi-con- 
jugate diameters are 45° and 135°. 

22. The polar of a point in a diameter is parallel to the 
conjugate diameter. 



178 ANALYTIC GEOMETRY. 

23. Find the equations of equal conjugate diameters. 

24. The length of a semi-diameter is I ; find the equation 
of the conjugate diameter. 

25. The angle between two equal conjugate diameters = 
60° ; find the eccentricity of the ellipse. 

26. Given a diameter, to construct the conjugate diameter. 

27. To construct two conjugate diameters which shall con- 
tain a given angle. 

28. To draw a tangent to a given ellipse parallel to a given 
straight line. 

29. Given an ellipse ; to find by construction the centre, 
foci, and axes. 

30. Find the rectangular equation of the ellipse, taking the 
origin at the left-hand vertex. 

31. Find the polar equation of an ellipse, taking as pole 
the left-hand focus. 

32. Find the polar equation of the ellipse, taking the centre 
as pole. 

33. Discuss the form of the ellipse by means of its polar 
equation. 

34. If the centre of an ellipse is the point (4, 7), and the 
major and minor axes are 14 and 8, find its equation, the axes 
being supposed parallel to the axes of co-ordinates. 

35. The equation of an ellipse, the origin being at the left- 
hand vertex, is 25 x 2 + 81 y 2 = 450 # ; find the axes. 

36. If the minor axis = 12, and the latus rectum = 5, 
what is the equation of the ellipse, the origin being taken at 
the left-hand vertex ? 



THE ELLIPSE. 179 

Find the centre and axes of the following ellipses : 

37. 4^ 2 + 3/ 2 + 8^-2y + l = 0. 

38. 9^ 2 + 16y 2 -36^-128y + 148 = 0. 

39. 4o; 2 + 363/ 2 +36y=:0. 

40. Find the eccentric angle cj> corresponding to the diam- 
eter whose length is 2 c. 

41. At the intersection of the ellipse b 2 x 2 + a 2 y 2 = a 2 b 2 and 
the circle x 2 -\-y 2 = ab tangents are drawn to both curves. 
Find the angle between them. 

42. How would you draw a normal to an ellipse from any 
point in the minor axis ? 

43. Find the equation of a chord bisected at a point (A, h). 

44. Prove that the length of a line drawn from the centre 
to a tangent, and parallel to either focal radius of the point 
of contact, is equal to the semi-major axis. 

45. A circle described on a focal radius will touch the 
auxiliary circle. 

46. Find the locus of the intersection of tangents drawn 
through the ends of conjugate diameters of an ellipse. 

47. Find the locus of the middle point of the chord joining 
the ends of two conjugate diameters. 

48. Find the locus of the vertex of a triangle whose base 
is the line joining the foci, and whose other sides are parallel 
to two conjugate diameters. 

49. Find the locus of the centre of a circle which passes 
through the point (0, 3) and touches internally the circle 
x 2 + y 2 = 25. 



CHAPTER VII. 

THE HYPERBOLA. 

Simple Properties of the Hyperbola. 

165, The Hyperbola is the locus of a point the difference of 
whose distances from two fixed points is constant. 

The fixed points are called the Foci, and a line joining any 
point of the curve to a focus is called a Focal Kadius. 

The constant difference is denoted by 2 a, and the distance 
between the foci by 2c. 

The fraction - is called the Eccentricity, and is denoted by 
a 

the letter e. Therefore c = ae. 

Since the difference of two sides of a triangle is always less 
than the third side, we must have in the hyperbola 

2a<2c, or a < c, or e>l. 

166. To construct an hyperbola, having given the foci, and 
the constant difference 2 a. 

I. By Motion (Fig. 66). Fasten one end of a ruler to one 
focus F f so that it can turn freely about F 1 . To the other 
end fasten a string. Make the length of the string less than 
that of the ruler by 2 a, and fasten the free end to the focus F. 
Press the string against the ruler by a pencil point P, and 
turn the ruler about F 1 . 

The point F will describe one branch of an hyperbola. 
The other branch may be described in the same way by inter- 
changing the fixed ends of the ruler and the string. 



THE HYPERBOLA. 



181 



II. By Points (Fig. 67). Let F, F 1 be the foci; then 

FF' = 2c 
Bisect FF' at 0, and from lay off OA = OA 1 = a. 
Then, AA' = 2a. 

AF f -AF = 2a + (c — a) — (c — a) = 2a. 

A'F -A'F'=2a + ic-a)--ic-a) = 2a. 
Therefore A and A 1 are points of the curve. 




Fig. 66. 



Fig. 67. 



In A A 1 produced mark any point D ; then describe two 
arcs, the first with F as centre and AD as radius, the second 
with F f as centre and A'D as radius; the intersections P, Q 
of these arcs are points of the curve. By merely interchang- 
ing the radii, two more points P, x?may be found. 

Proceed in this way till a sufficient number of points has 
been obtained ; then draw a smooth curve through them. 

Through draw £P f JL to FF f ; since the difference of 
the distances of every point in the line BB ! from the foci is 
0, therefore the curve cannot cut the line BB f . 

The locus evidently consists of two entirely distinct parts or 
branches, symmetrically placed with respect to the line BB 1 . 



182 



ANALYTIC GEOMETRY. 



167. The point 0, half way between the foci, is the Centre. 

The line A A 1 passing through the foci and limited by the 
curve is the Transverse Axis. 

The points A, A 1 , where the transverse axis meets the 
curve, are called the Vertices. 

The transverse axis is equal to the constant difference 2 a, 
and is bisected by the centre (§ 166). 




Fig. 68. 

The line BB f passing through perpendicular to A A 1 
does not meet the curve (§ 166) ; but if B, B' are two points 
whose distances from the two vertices A, A* are each equal to 
c, then BB f is called the Conjugate Axis, and is denoted by 2b. 

Since A AOB = AAOB\ OB = OB'^b; that is, the con- 
jugate axis is bisected by the centre. 

In the triangle AOB, OA = a, OB = b, AB = c ; hence 

# = '#+&« 

The chord passing through either focus perpendicular to 
the transverse axis is called the Latus Eectum, or Parameter. 

Note. Since a and b are equal to the legs of a right triangle, a 
may be either greater or less than b\ hence the terms "major" and 
"minor" are not appropriate in the hyperbola. 



THE HYPERBOLA. 183 

168. To find the equation of the hyperbola, having given the 
foci, and the constant difference 2 a, 

By proceeding as in the case of the ellipse (§ 131), and sub- 
stituting b 2 for c 2 — a 2 , we obtain 

S-S = 1 ' [40] 

The lengths r, r 1 of the focal radii for any point (x, y) are 

r = ex — a and r f = ex + a. ' 
The equations of the ellipse and the hyperbola differ only 
in the sign of b 2 . The equation of the hyperbola is obtained 
from that of the ellipse by changing -\-b 2 to — b 2 . In general 

Any formula deduced from the equation of the ellipse is 
changed to the corresponding formula for the hyperbola by 
merely changing -\-b 2 to — b 2 , or b to b V— 1. 

169. A discussion of equation [40] leads to the following 
conclusions : 

(i.) The curve cuts the axis of x at the two real points 
- (a, 0) and (—a, 0). 

(ii.) The curve cuts the axis of y at the two imaginary 
points (0, &V-1) and (0, - bs/^V). 

(iii.) No part of the curve lies between the straight lines 
x = -\-a and x = — ■ a. 

(iv.) Outside these lines the curve extends without limit 
both to the right and to the left. 

(v.) The greater the abscissa, the greater the ordinate. 

(vi.) The curve is symmetrical with respect to the axis of x. 

(vii.) The curve is symmetrical with respect to the axis of y. 

(viii.) Every chord which passes through the centre is bi- 
sected by the centre. This explains why the point half way 
between the foci is called the centre. 

The two distinct parts of the curve are called the right-hand 
and the left-hand branches. 



184 



ANALYTIC GEOMETRY. 



170. An hyperbola whose transverse and conjugate axes 
are equal is called an Equilateral Hyperbola. Its equation is 

[41] 



y* = a* 



The equilateral hyperbola bears to the general hyperbola 
the same relation that the auxiliary circle bears to the ellipse. 




Fig. 6 9 . 



171. The hyperbola which has BB' for transverse axis, and 
A A' for conjugate axis, obviously holds the same relation to 
the axis of y that the hyperbola which has A A 1 for transverse 
axis and BB f for conjugate axis holds to the axis of x. 

Therefore its equation is found by simply changing the 
signs of a 2 and b 2 in [40], and is 



a 2 U 2 



xT 
or -j- 

a 1 



b 1 



-1. 



The two hyperbolas are said to be Conjugate. 



THE HYPERBOLA. 185 



172. The straight line y == mx, passing through the centre 

x 2 v 2 
of the hyperbola — 2 — j- 2 = 1, meets the curve in two points, 



a' 

the abscissas of which are 



+ a b _ — ah 



V& 2 — mV V& 2 — m 2 a 2 

Hence the points will be real, imaginary, or situated at 
infinity, as b 2 — m 2 a 2 is positive, negative, or zero ; that is, as 

m is less than, greater than, or equal to -• 

a 

The same line, y = mx, will meet the conjugate hyperbola 

x 2 y 2 

-~2 — j- 2 = — 1 in two points, whose abscissas are 

+ ah — ab 



^Jm 2 d l — b' 1 Vm 2 a 2 — 6 2 

Hence these points will be imaginary, real, or situated at 

infinity, as m is &ss than, greater than, or equal to — 

a 

Whence 

7f a straight line through the centre meet an hyperbola in 
imaginary points, it will meet the conjugate hyperbola in real 
points, and vice versa, 

173. An Asymptote is a straight line which passes through 
finite points, and meets a curve in two points at infinity. 
We see from § 172 that the hyperbola 



x 2 



,/ 2 



A2 x 



has two real asymptotes passing through the centre of the 

curve, and having for their equations y = -f - x and y = x ; 

or, expressed in one equation, 

/y»2 -»/2 

S -! = <>• [42] 



186 ANALYTIC GEOMETRY. 

Ex. 36. 

What is the equation of an hyperbola, if 

1. Transverse axis = 16, conjugate axis = 14 ? 

2. Conjugate axis = 12, distance between foci = 13 ? 

3. Distance between foci = twice the transverse axis ? 

4. Transverse axis = 8, one point = (10, 25) ? 

5. Distance between foci = 2 c, eccentricity = V2 ? 

6. Prove that the latus rectum of an hyperbola is equal to 
2tf 

a 

7. The equation of an hyperbola is 9 a? 2 — 16 y 2 = 25 ; find 
the axes, distance between the foci, eccentricity, and latus 
rectum. 

8. Write the equation of the hyperbola conjugate to the 
hyperbola 9a? 2 — 16y 2 = 25, and find its axes, distance between 
its foci, and its latus rectum. 

9. If the vertex of an hyperbola bisects the distance from 
the centre to the focus, find the ratio of its axes. 

10. Prove that the point (a?, y) is without, on, or within the 

a? ij 
hyperbola, according as — — ?- — 1 is negative, zero, or posi- 
tive. 

11. Find the eccentricity of an equilateral hyperbola. 

12. The distance of any point of an equilateral hyperbola 
from the centre is a mean proportional between its focal radii. 

13. The asymptotes of an hyperbola are the diagonals of 
the rectangle CDEF (Fig. 69, p. 184). 

14. Find the foci and the asymptotes of the hyperbola 
16a? 2 -9y 2 = 144. 



THE HYPERBOLA. 187 

15. The asymptotes of an equilateral hyperbola are perpen- 
dicular to each other. Hence the equilateral hyperbola is 
also called the rectangular hyperbola. 

16. An hyperbola and its conjugate have the same asymp- 
totes. 

17. Find the length of a perpendicular dropped from the 
focus to an asymptote. 

Tangents and Normals. 

Note. The results stated in the following six sections are established 
in the same way as the corresponding propositions relating to tangents 
and normals to an ellipse. We shall, therefore, omit the proofs. 

174. The equation of the tangent at (x Y) 3/1) is 

^_^/_l [431 

175. The equation of the normal at (x 1% y x ) is 

b* a a 2 b 2 * L J 

176. The subtangent = -> the subnormal = — =^- 

& x 1 a 2 

177. The tangent and the normal at any point of an hyper- 
bola bisect the angles formed by the focal radii of the point 
(§ 144). 



178. The straight line whose equation is y = ?nx ± -\m l d l — b 2 
is a tangent for all values of m (§ 146). 

179. The equation of the director circle of an hyperbola is 
z 2 + f = a 2 - b 2 (§ 147). 



188 x^NALYTIC GEOMETRY. 

Ex. 37. 

1. Find the equations of tangent and normal to the hyper- 
bola IQx 2 — 9y 2 = 112 at the point of contact (4,4). Also 
find the lengths of the subtangent and the subnormal. 

2. Show that in an equilateral hyperbola the subnormal is 
equal to the abscissa of the point of contact. 

3. The equations of the tangent and the normal at a point 
of an equilateral hyperbola are 5x — 4y = 9, 4.r + 9y — 56. 
What is the equation of the hyperbola, and what are the 
co-ordinates of the point of contact ? 

4. For what points of an hyperbola is the subtangent equal 
to the subnormal ? 

5. To draw a tangent and a normal to an hyperbola at a 
given point of the curve. 

6. If an ellipse and an hyperbola have the same foci, prove 
that the tangents to the two curves drawn at their points of 
intersection are perpendicular to each other. 

7. Prove that the asymptotes of an hyperbola are tangent 
to it at infinity. 

8. Prove that the length of a normal in an equilateral 
hyperbola is equal to the distance of the point of contact from 
the centre. 

9. Find the distance from the origin to the tangent through 
the end of the latus rectum of the equilateral hyperbola 
x 2 — y 2 = a 2 . 

10. AYhat condition must be satisfied in order that the 

x v x 2 v 

straight line - — \- — = 1 may touch the hyperbola —, — t= =1? 

11. When is the director circle of an hyperbola imaginary ? 

12. Find the locus of the foot of a perpendicular dropped 
from the focus of an hyperbola to a tangent. 



THE HYPERBOLA. 189 

Ex. 38. (Review.) 

1. The ordinate through the focus of an hyperbola, pro- 
duced, cuts the asymptotes in P and Q. Find PQ and the 
distances of P and Q from the centre. 

2. In the hyperbola 9x 2 — 16y 2 = 144 what are the focal 
radii of the points whose common abscissa is 8 ? What other 
points have equal focal radii ? 

3. What relation exists between the sum of the focal radii 
of a point of an hyperbola and the abscissa of the point ? 

4. Prove that in the equilateral hyperbola every ordinate 
is a mean proportional between the distances of its foot from 
the vertices of the curve. Hence find a method of construct- 
ing an equilateral hyperbola when the axes are given. 

5. In the equilateral hyperbola the distance of a point from 
the centre is a mean proportional between its focal radii. 

6. In the equilateral hyperbola the bisectors of the angles 
formed by lines drawn from the vertices to any point of the 
curve are parallel to the asymptotes. 

7. If e, e f are the eccentricities of two conjugate hyperbolas, 

e 1 e' 2 

8. Through the positive vertex of an hyperbola a tangent is 
drawn. In what points does it cut the conjugate hyperbola ? 

9. The sum of the reciprocals of two focal chords perpen- 
dicular to each other is constant. 

10. Through the foot of the ordinate of a point in an equi- 
lateral hyperbola a tangent is drawn to the circle described 
upon the transverse axis as diameter. What relation exists 
between the lengths of this tangent and the ordinate of the 
point ? 



190 ANALYTIC GEOMETRY. 

11. In an equilateral hyperbola find the equations of tan- 
gents drawn from the positive end of the conjugate axis. 

12. From what point in the conjugate axis of an hyperbola 
must tangents be drawn in order that they may be perpen- 
dicular to each other ? 

13. "What condition must be satisfied that a square may be 
constructed whose sides shall be parallel to the axes of an 
hyperbola and whose vertices shall lie in the curve ? 

14. Find the equation of the chord of the hyperbola 
16x 2 — 9y 2 = 144 which is bisected at the point (12,3). 

15. Find the equation of a tangent to the hyperbola 
I6x 2 — 9?/ 2 = 144 parallel to the line y=Ax-3. 

16. Determine the points in an hyperbola for which the 
length of the tangent is twice that of the normal. 

17. A chord of an hyperbola which touches the conjugate 
hyperbola is bisected at the point of contact. 



SUPPLEMENTARY PROPOSITIONS. 

Note. Many of the following propositions are closely analogous to 
propositions already established for the ellipse ; hence the proofs are 
omitted, and references given to the chapter on the ellipse. 

180. Two tangents can be drawn to an hyperbola from any 
point (h, k) ; and they will be real, coincident, or imaginary, 
as the point is without, on, or within the curve (§ 148). 

h 2 Jc 2 

The two tangents will be real if 1 is negative. 

S a 2 y 2 6 

Likewise two real tangents can be drawn from (h, Jc) to the 

h 2 Jc 2 
conjugate hyperbola if — — — + 1 is negative. 
a o 



THE HYPERBOLA. 191 

h 2 k 2 
Hence it follows that if — — — has any value between 
a 1 b 2 J 

and — 2, a pair of real tangents can be drawn from (A, h) to 
each* hyperbola. 

181. The straight line passing through the points of contact 
of the two tangents drawn to an hyperbola from any point P 
is called the Polar of P with respect to the hyperbola ; and P 
is called the Pole of this straight line. 

The polars of the foci are called the Directrices. 
The equation of the polar of the point (h, h) is 

^_| = l. (U49) 

The propositions in §§ 80-82 hold true for poles and polars 
with respect to an hyperbola, and may be proved in the same 
way. 

182. The locus of the middle points of chords parallel to 
the line y = mx is -, 2 

y = 4 £ - (§ 152 ) 

a l m 

This locus is called a Diameter of the hyperbola. 
Every diameter passes through the centre. 
The chords bisected by a diameter are called the Ordinates 
of the diameter. 

183. If m! is the slope of the diameter, bisecting chords 
parallel to the line y = mx, then 

mm' = g; [45] 

and from the symmetry of this equation we infer that 

Tf one diameter bisects all chords parallel to another \ the 
second diameter will bisect all chords parallel to the first. 

Two diameters draw T n so that each bisects all chords par- 
allel to the other are called Conjugate Diameters (§ 153). 



192 ANALYTIC GEOMETRY. 

184. Since the product of the slopes of two conjugate diam- 
eters is 2 

mrw == — > 
dr 

it follows that m and m) must agree in sign ; therefore 
Two conjugate diameters lie in the same quadrant. 

Also, if m in absolute magnitude is less than -> then m ! 
, a 

must be greater than — But the slope of the asymptotes is 

b a 

equal to ±- Therefore 
a 

Two conjugate diameters lie on opposite sides of the asymp- 
tote in the same quadrant ; and of two conjugate diameters, one 
meets the curve in real points and the other in imaginary points 
(§ 172). 

185. The length of a diameter which meets the hyperbola 
in real points is the length of the chord between these points. 

If a diameter meets the hyperbola in imaginary points, 
that is, does not meet it at all, it will meet the conjugate 
hyperbola in real points (§ 172) ; and its length is the length 
of the chord between these points. 

A comparison of the equations of two conjugate hyper- 
bolas will show that if a diameter meet one of the hyper- 
bolas in the imaginary point (AV— 1, 7jV— 1), it will meet 
the other in the real point (h, k) ;' hence the length of the 
semi-diameter will be V/i 2 -f- k'\ 

186. The equations of an hyperbola and its conjugate differ 
only in the signs of a 2 and b 2 . But this interchange of signs 
does not effect the equation 

b 2 
mm* == — , Therefore 
a 2 

If two diameters are conjugate with respect to one of two con- 
jugate hyperbolas, they will be conjugate with respect to the other. 



THE HYPERBOLA. 



193 



Thus, let POP and QOQ 1 (Fig. 70) be two conjugate di- 
ameters. Then POP bisects all chords parallel to QOQ 1 that 
lie within the branches of the original hyperbola and between 
the branches of the conjugate hyperbola ; and QOQ bisects all 
chords parallel to POP that lie within the branches of the 
conjugate hyperbola and between the branches of the original 
hyperbola. 



Ds 




Fig. 70. 



From the above theorem it follows immediately that 

If a straight line meet each of two conjugate hyperbolas in 
two real points, the two portions of the line contained between 
the hyperbolas are equal {thus, BD = B'D\ Fig. 70). 

187. The tangent drawn through the end of a diameter is 
parallel to the conjugate diameter (§ 154). 



194 ANALYTIC GEOMETRY. 

188. Having given the end {x x , y^) of a diameter, to find the 
end (x 2 , y 2 ) of the conjugate diameter. 

If (x lf 2/i) is a point of the hyperbola 

i, a) 



xf_ _y^ 
a 2 V ' 



we know (§ 172) that (x 2 , Vi) will be a point of the conjugate 

hyperbola 2 2 

— — ^- = — 1. (2) 

a 2 b 2 K J 

If the equation of the diameter through (x lf y^ be 

y = m#, 

then m = — 

x x 

If the equation of the diameter through (x 2 , y 2 ) be 

y — m'^, 

then m'=-%- (§ 182). 

Hence the equation y = m f x may be written 

»-^- (3 > 

The values of x 2 and y 2 are now found by solving equations 
(2) and (3), and are 

a b 

o a 

The positive signs belong to one end, and the negative signs 
to the other end, of the conjugate diameter. 

189. If 6 denote the angle formed by two conjugate semi- 
diameters, a' and V , their lengths, then sin = — — (§ 158). 



THE HYPERBOLA. 195 

190. To find the equation of an hyperbola referred to any 
pair of conjugate diameters as axes of co- or d mates. 

If a', b' denote the two semi-diameters, the required equa- 
tion is 2 2 

a 12 V 2 K } 

The method of solving the problem is the same as that used 
in § 160 ; but since the intercept of the curve on the axis of y 
is imaginary, § 167, the sign of B in § 160 will be negative. 

Since the form of equation (1) is the same as that of the 
equation referred to the axes of the curve, it follows that all 
formulas which have been obtained without assuming the axes 
of co-ordinates to be at right angles to each other hold good 
when the axes of co-ordinates are any two conjugate diam- 
eters. For example, the equation of the asymptotes of the 
hyperbola represented by equation (1) is 

^_l! = 
a' 2 V 2 

191. The tangents through the ends of two conjugate diam- 
eters meet in the asymptotes. 

Let the ends of the diameters be the points (x lt y^) and 

(#2, 3/2) ; then the equations of tangents through (x\, y^) and 

2 , y 2 ) will be x^x_ _ yy_ m 

a 2 b 2 ' V J 

and ^_Sg? = _ L (2) 

a 2 b 2 v J 

The asymptote y = -x meets the tangent represented by 
a 

(1) in the point / a ^ a tf \ 

\bx 1 —ay 1 f bx 1 —ayj 
and the tangent represented by (2) in the point 

/ a 2 b ab 2 

\ay 2 — bx 2 ay 2 —bx 2i 



196 ANALYTIC GEOMETRY. 

But from § 188, bx 2 = ay x and ay 2 = bx 1 . 

Therefore ay 2 — bx 2 — bx x — ay x . 

Hence we see that the two points of meeting coincide. 

192. To find under what condition an equation will repre- 
sent an hyperbola when of the form 

Ax 2 +By 2 +Bx + Uy+F=0. 

If neither A nor B is zero, the equation may be written 

J . D\\ p / , B\ 2 B 2 . E 2 „ 

By proceeding as in § 161, we find that the equation repre- 
sents in general an hyperbola if A and B are neither of them 
zero, and have unlike signs. 

The axes of the hyperbola are parallel to the axes of x and y, 

(T) ~F\ 

— — — ' — — — )• 
2 A 2 BJ 

193. If a straight line cut an hyperbola and its asymptotes 
the portions of the line intercepted between the curve and its 
asymptotes are equal. 

Let CC (Fig. 71) be the line meeting the asymptotes in C, C 
and the curve in B, B 1 , and let the equation of the line be 

y — mx-\-c. (1) 

Let M be the middle point of the chord BB' ; then (§ 182) 
the equation of the diameter through m is 

y=¥- (2) 

a m 

By combining equation (1) with the equations of the asymp- 
totes, we obtain the co-ordinates of the points Cand C f ; taking 



THE HYPERBOLA. 



197 



the half-sum of these values, we get for the co-ordinates of the 
point half way between C and C the values 

b 2 c 



b 2 — m 2 a 2 



y= 



Bs 



tM 



(r 



P'h 



S 



\B r 



ic 



'By 



Fig. 71. 



These values satisfy equation (2) ; therefore the point half 
way between C and C f coincides with M ; therefore MC—MCK 
And since MB = MB\ therefore £C=JB'C. 

Let CC be moved parallel to itself till it becomes a tangent 
at P, meeting the asymptotes in H, S ; then the points J3, J3' 
coincide at P, and w r e have PjR — jPS. Hence 

The portion of a tangent intercepted by the asymptotes is 
bisected by the point of contact 



198 



ANALYTIC GEOMETRY. 



194. To find the equation of an hyperbola referred to the 
asymptotes as axes of co-ordinates. 




Fig. 72. 

Let the lines OB, 00 (Fig. 72) be the asymptotes, A the 
vertex of the curve, and let the angle AOC= a. 

Let the co-ordinates of any point P of the curve be x, y 
when referred to the axes of the curve, and x\ y 1 when referred 
to OB, 00 as axes of co-ordinates. 

Draw PM± to OA, PNW to 00; then 

x = OM, y = 3£P, x' = ON, y' = NP. 

X = ON COS a + NP COS a = (V -f ?/ ) COS a, 
y = NP sin a — ON sin a = (y 1 — x') sin a. 

Hence, by substitution [40], we obtain 

(V + y') 2 cos 2 a (y* — x') 2 sin 2 a 



But from the relation tan a = 
b 2 



b 2 
b 



= 1. 



we have 



sin a - 



COS' a = ■ 



a 2 + 6 2 a 2 + b 2 

Substituting these values, and dropping accents, we have 

±xy = a 2 + 6 2 . [46] 



THE HYPERBOLA. 199 

195. The following method of showing that an hyperbola 
has asymptotes, and finding their equations, is more general 
than the method given in §§ 172, 173. 

The abscissas of the points where the straight line y = mx-\-c 
meets an hyperbola are found by solving the equation 



X 2 


(mx -f- c) 2 i 


b 2 - 


b 2 l ' 

- m 2 a 2 o 2 mc 
xr — X 



o ~\~ o /i \ 

= — • <*> 

This equation is identical with the equation 

Ax 2 + 2Bx+C=0 (2) 

if A = b *- m * a \ B = -™, C=-^±?. 

a l b 2 b 2 b 2 

If x u x 2 are the roots of (2), 

-£ + -V.B 2 -AC_ C 



#i = 



-B-^B'-AC 



-B-^B 2 -AC C 

Z*= : 



* -b + Vb 2 -ac 

c 

If A = and also ,5 = 0, w T e have x\ = %» = — = oo : and 



the line y = mx + c will meet the curve in two points at 

infinity. 

If A = 0, m = ± -• If B = 0, c = 0. Therefore there are 

, a . b b 
two asymptotes; their equations are y = -x and y = x. 

If only ^4 = 0, then m = dz -» the straight line is parallel 
to an asymptote, and 

x t = = ^— » # 2 = °° • Therefore 

2B zpic 

A straight line parallel to an asymptote meets the curve in 
one finite point and in one point at infinity. 



200 



ANALYTIC GEOMETRY. 



196. To find the locus of a point which moves so that its 
distance from a fixed point bears to its distance from a fixed 
straight line a constant ratio greater than unity. 




Fig. 73. 



Let e — the constant ratio, and 2p = the distance from the 

fixed point F (Fig. 73) to the fixed line DN Choosing for 

axes the line drawn through F perpendicular to DN, and the 

fixed line DN, we obtain the same equation as that found in 

§ 162, 

(1 - e 2 )x 2 + y 2 - ±pz + 4p 2 = 0. (1) 

But since e is greater than 1, the equation now will repre- 
sent an hyperbola (§ 195), the centre of which is at the 
2p 



(2) 



point \TZr^ ° 

Transferring the origin to the centre 0, we get 

a - £ 2 v a , i 



2ep 



x 2 - 



(2ep) 



y=i- 



THE H1TERB0LA. 201 

Putting a = ^ — ^ b = ^Jl — (2) becomes 
i — e we 2 — 1 

the equation of an hyperbola in the ordinary form. 

It may be shown, as in § 162, that the fixed point F coin- 
cides with the focus of the curve, and that the constant ratio 
e is equal to the eccentricity. 

This locus is often taken as the definition of an hyperbola, 
F being called the Focus, and the fixed line DN the Directrix. 

The symmetry of the curve with respect to the conjugate 
axis shows that there are two foci and two directrices sym- 
metrically placed with respect to the conjugate axis. 

197. To find the polar equation of an hyperbola, the left- 
hand focus being take?i as pole. 

For a point in the right-hand branch, the distance to the 
pole, which is the remote focus, is (§168) 

p = ex -{- a, 

the distance x being reckoned from the centre. 

Now x — p cos — c = p cos 6 — ae. 

Whence, by substitution and reduction, 

a(e* — 1) 



e cos 6 



[47] 



Ex. 39. 

1. What is the polar of the point (— 9, 7) with respect to 
the hyperbola 7^ - 12y 2 = 112? 

2. Find the equations of the directrices of an hyperbola. 

3. Find the angle formed by a focal chord and the line 
which joins its pole to the focus. 



202 ANALYTIC GEOMETRY. 

4. Find the pole of the line Ax-\-By-\- C with respect to 
an hyperbola. 

5. Find the polar of the right-hand vertex of an hyperbola 
with respect to the conjugate hyperbola. 

6. Find the distance from the centre of an hyperbola to 
the point where the directrix cuts the asymptote. 

7. If {x Xl i/i) and (x 2) y 2 ) are the ends of two conjugate 
diameters, then x x v ? , 

8. The equation of a diameter in the hyperbola 25 # 2 — 16y 2 
= 400 is 3y = x. Find the equation of the conjugate diam- 
eter. 

9. In the hyperbola 49 # 2 -- 4y 2 == 196, find the equation of 
that chord which is bisected at the point (5, 3). 

10. Find the length of the semi-diameter conjugate to the 
diameter y = Sx in the hyperbola 9x 2 — 4y 2 = 36. 

11. The area of the parallelogram formed by drawing tan- 
gents through the ends of two conjugate diameters is constant, 
and equal to 4a&. 

12. If a\ b' are the lengths of two conjugate semi-diameters, 
then a i2 _ v % ^ a 2 __ j2 

13. Prove that PQ (Fig. 70) is parallel to one asymptote 
and bisected by the other. 

14. An asymptote is its own conjugate diameter. 

15. The conjugate diameters of an equilateral hyperbola 
are equal. 

16. Having given two conjugate diameters in length and 
position, to find by construction the asymptotes and the axes. 

17. To draw a tangent to an hyperbola from a given point. 



THE HYPERBOLA. 203 

18. Find the equation of a conjugate hyperbola referred to 
its asymptotes as axes. 

19. Find the equation of a tangent at any point (x h y^) of 
the hyperbola 4.ry = a 2 -f- b 2 . 

20. Find the equation of an hyperbola, taking as the axis 

y (i.) the tangent through the left-hand vertex ; 
(ii.) the tangent through the right-hand vertex. 

21. Trace the form of an hyperbola by means of the polar 
equation, p. 201. 

22. Find the polar equation of an hyperbola, taking the 
right-hand focus as pole. 

23. Find the polar equation of an hyperbola, taking the 
centre as pole. 

24. Show that the equation 

x 2 -if — 2x-4;y + l = 
represents an hyperbola. Find its centre and axes, and con- 
struct roughly the curve. 

25. The distance from a fixed point to a fixed straight line 
is 10. Find the locus of a point which moves so that its dis- 
tance from the fixed point is always twice its distance from 
the fixed line. 

26. Through the point (— 4, 7) a straight line is drawn to 
meet the axes of co-ordinates, and then revolved about this 
point. Find the locus of its middle point. 

27. A straight line has its ends in two fixed perpendicular 
lines, and forms with them a triangle of constant area a 2 . 
Find the locus of the middle point of the line (see § 000). 

28. The base a of a triangle is fixed in length and position, 
and the vertex so moves that one of the base angles is always 
double the other. Find the locus of the vertex. 



CHAPTER VIII. 
LOCI OF THE SECOND ORDER. 

198. The locus represented by an equation of the second 
degree is called a Locus of the Second Order. 

We have seen, in the preceding chapters, that the circle, 
parabola, ellipse, and hyperbola are loci of the second order. 
We now propose to inquire whether there are other loci of 
the second order besides the four curves just named; in other 
words, to find what loci may be represented by equations of 
the second degree. 

For this purpose we shall write the general equation of the 
second degree in the form 

Ax 2 + By 2 + Cxy + Dx + Ey + F=0, (1) 

and shall assume that the axes of co-ordinates are rectangular. 
This assumption will in nowise diminish the generality of 
our conclusions ; for if the axes be supposed oblique, we can 
change them to rectangular axes, and this change will not 
alter the degree of the equation or the nature of the locus 
which it represents (§ 100). 

199. If we suppose the coefficients of equation (1) to be 
susceptible of all values including zero, we see that (1) in- 
cludes the general equation of the first degree as a special 
case when A = B — C= 0. 

But even if no one of the coefficients is zero, they may 
stand in such a relation to one another that the equation can 
be resolved into two linear factors, and therefore represents 
straight lines, real or imaginary. 



LOCI OF THE SECOND ORDER. 205 

In order to find what this relation is, let us solve (1) with 
respect to one of the variables. Choosing y for this purpose, 

we obtain Cx + E 1 — 

y = -—^~^^^Lx' + Mx+N 1 (2) 

where 

L = C*-iAB i M=2(CE-2BD), N=E 2 -±BF. 

If Lx 2 -j- Mx-\-N be the square of a binomial of the form 
Sx+T, then the value of y may be written 

Cx + E Sx + T 



y 



2B ~ 2B 



and the locus of (1) will in general be a pair of straight lines. 
Now, from Algebra, we know that the condition that 
Lx 2 + Mx + N should be a perfect square is 

_af 2 -4ZiY=0; 
or, substituting the values of B, M, and iV", 

(CE-2BD) 2 - (C 2 - 4tAB) (E 2 -±BF) = 0, 

or E(C 2 -4.AB)+AE 2 +BE> 2 -CBE^0. (3) 

The quantity on the left-hand side of equation (3) is usually 
denoted by A, and is called the Discriminant of equation (1). 
And w T e may conclude that (1) represents straight lines (real 
or imaginary) whenever A — 0. 

200. In order to simplify the form of equation (1), let us 
change the origin to the point (A, k), and then so choose the 
values of h and h that the terms involving the first powers of 
v and y will vanish. Making the change by substituting in 
(1) a?+ h for x, and y + Jc for y, we find that the coefficients 
A, B, and C remain unaltered, and we may write the trans- 
formed equation 

Ax 2 + By 2 + Cxy + B'x -\-Ehj = R (4) 



x 



206 ANALYTIC GEOMETRY. 

where . D' = 2Ah + Ck + D, 
E f = 2Bk + Ch+E, 
B = -[Ah 2 +Bk 2 + Chk + Bh + Ek+E]. 

The values of A and lc which will make I) 1 and E' vanish 
are evidently found by solving the equations 

2Ah+CJc+I) = 7 
2Bk + Ch + E^0, 

' , CE-2BD 7 CD-2AE 

and are A = — -, re = — -• 

4:AB—C 2 4:AB—C 2 

The value of B can now be reduced to a form very easily 
remembered : 

R = - [ JLA 2 + DF + CM + DA + D£ + F] 

= --i[(2Ah+Ck+B)h+(2BJc+Ch+E)k+Bh+Ek+2F] 
= - i(Z>'A + £"A + DA + Eh + 2F) 
= -i(Dh + Ek+2F) 

1 2 BE 2 - CDE+ 2AE 2 - CDE+ 2F(C 2 - 4AB) 

2 C 2 -4AB 
= A 

where 2 = 4^4i?-C 2 . 

Equation (4) may now be written 

Ax 2 +By 2 +Cxy=B. (5) 

From the form of (5) we see that if (x, y) be a point of the 
locus, so also is (— x, — y) ; that is, the new origin is a point 
so placed that it bisects every chord passing through it. A 
point having this property is called a Centre of a locus of 
the second order. 

The values of A and h are evidently single ; hence a locus 
of the second order cannot have more than one centre. 

The values of A and h are finite, provided 2 or 4^4 B — C 2 
is not zero. If, however, 2 = 0, the values of A and h become 









LOCI OF THE SECOND ORDER. 207 

infinite or indeterminate. In this case a change of origin to 
the centre is obviously impossible, and a different method of 
reduction must be found. 

Hence it will be convenient to divide loci of the second 
order into two classes : those which have a finite centre, and 
those which do not. The ellipse and the hyperbola belong 
to the first class ; the parabola, to the second class. 

The class to which the locus of a given equation belongs 
is ascertained by seeing whether the value of 2, namely, 
4:AB — C 2 , is or is not equal to zero; on this account 2 
may be called the Criterion of the general equation of the 
second degree. 

CLASS I. 2 NOT ZERO. 

201. Equation (5) is the general equation of loci of the 
second order which have a centre, referred to the centre as 
the origin of co-ordinates. 

If in equation (5) we place x = 0, we obtain two values of 
y equal in magnitude and opposite in sign. Since the axis of 
y is not limited as to direction, we infer that every chord pass- 
ing through the centre is bisected at the centre. Hence a chord 
passing through the centre is called a Diameter. 

We can get rid of the term involving xy by another change 
of axes. For this purpose we must change the direction of 
the axes through an angle 0, keeping the origin unaltered, 
and then determine the value of 6 by putting the new term 
which involves xy equal to zero. 

The change is made by substituting for x and y, in equation 
(5), the respective values (§ 95), 

x cos — y sin 0, 
x sin 6 -f- y cos ; 

and equation (5) now becomes 

Px 2 + Qy 2 + C'xy == R % 



208 ANALYTIC GEOMETRY. 

where P = A cos 2 6 + B sin 2 6 + C sin cos 0, (6) 

Q = A sin 2 +B cos 2 - Csin6 cos6>, (7) 

C' = 2(B-A)sm0cos6+C(cos 2 6-sm 2 6). (8) 

Putting C = 0, we obtain 

(A -B) sint<9 - Cco§6 = 0, (9) 

or tan 2(9 = —-^—, (10) 

A — B 

a relation which will always give real values for 0, since the 
tangent of an angle may have any value, positive or negative. 

Since the angles which correspond to a given value of a 
tangent differ by 180°, the values of obtained from (10) will 
differ by 90° ; hence the new axes determined by (10) are 
limited to a single definite pair of perpendicular lines, passing 
thro ugh the centre. 

The coefficients P, Q, A, B, C are connected by simple 
relations, which are ^dependent oJAthe value of 0, and which 
may be found as follows. 

From (6) and (7), by addition and subtraction, 

P+Q = A+B, (11) 

P—Q = (A—B) cosie + CslrZO. (12) 

Equation (9) may be written 

= (A-£)sirle-Ccos£e. (13) 

Adding the squares of (12) and (13), we have 

(p - Qy = (A -sy + c\ (14) 

P-Q = ±V(A-Bf + 0\ (15) 

Whence, from (11) and (15), 

P^j[A + P±V (A-£y + C r l (16) 

Q = i [A + B =f V(^ - By + C 1 ]. (17) 

These values of P and Q are always real. 



LOCI OF THE SECOND ORDER. 209 

Finally, by squaring (11) and subtracting (14), we obtain 

±PQ = ±AB-C 2 = X (18) 

In applying these formulas the question arises which sign 
should be chosen before the radical in equation (15). If we 
take for 26 the smallest positive angle which corresponds to 
the value of tan 26 in (10), then the value of 26 must lie 
between 0° and 180°, and sin 26 must be positive. If, now T , 
in equation (12) we substitute for cos 26, its value, obtained 
from (9), equation (12) will become 

O 

The form of this equation shows that P — Q must have the 
same sign as that of C. 

Equation (5) is now T reduced to the simple form 

Px 2 + Qif=B. (19) 

We see from (19) that the axes of co-ordinates are now 
so placed that each axis bisects all chords parallel to the 
other axis. Two lines so drawm through the centre of a curve 
of the second order that they have this property are called 
Conjugate Diameters. 

From w T hat proceeds we may infer that in those loci of the 
second order which have a centre there exists one, and only 
one, pair of rectangular conjugate diameters. 

These two diameters are called the Axes of the curve. Hence 
equation (19) is the general equation of curves of the second 
order, referred to the centre as origin and to their axes as the 
axes of co-ordinates. 

202, The nature of the locus represented by equation (19) 
depends upon the signs of P, Q, and R. There are two 
groups of cases, according as S is positive or negative, and 
three cases in each group. 



210 ANALYTIC GEOMETRY. 

Group 1. 2 Positive. 

Then equation (18) shows that P and Q must agree in sign. 
(A and B must also agree in sign.) 

Case 1. If P agree in sign with P and Q, then (§ 135) 
the locus is an ellipse, having for semiaxes 

HI »H! 

Case 2. If i? differ from P and Q in sign, no real values 
of x and y w T ill satisfy (19), so that no real locus exists. But 
it is usual to say in this case that the locus is an imaginary 
ellipse. 

Case 3. If P = 0, the locus is a single point, namely, the 
origin. 

Group 2. S Negative. 

Then (18) shows that P and Q must have unlike signs. 

Case 1. If P agrees in sign with P, we may, by division 
(and changing the signs of all the terms if necessary), put 
equation (19) into the form of equation [40], page 183. 
Therefore the locus is an hyperbola, with its transverse axis 
on the axis of x, and having for semiaxes 

Case 2. If P agrees in sign with Q, we may, by division 
(and change of signs if necessary), put equation (19) into the 
form of equation (2), page 184. Therefore the locus is an 
hyperbola, with its transverse axis on the axis of y. 

Case 3. If P — 0, the locus consists of two straight lines, 
intersecting at the origin, and having for their equations 



2/ = ±\J 



P 

— ^x. 

-Q 



LOCI OF THE SECOND ORDER. 211 



CLASS II. 2 = 0. 



203. Let us now suppose that, in the general equation, 2, 
or 4 AB — C 2 , is equal to zero. When this is the case, A and 
B must have like signs, and we shall assume that A and B 
are positive ; if they happen to be negative, we may make 
them positive by multiplying the equation by the factor — 1. 

The existence of a curve of this class is immediately shown 
by the form of the equation ; for the condition 4 AB — C 2 = 
is also the condition that the first three terms, Ax 2 -[-By 2 -\-Cxy, 
form a complete square. 

Throughout this section we shall also assume that C is 
not zero ; this being assumed, it follows from the relation 
4 AB — C 2 = that neither A nor B can be zero. 

When 2 = 0, the values of the co-ordinates of the centre, 

, CE-2BD 7 CD-2AE 
4cAB-0 2 4:AB-C 2 

become indeterminate or infiyiite, according as the numerators 
of the values are, or are not, equal to zero. In both cases a 
change of origin to the centre (as in § 200) is impossible. 

In the case where h and h are indeterminate, however, the 
nature of the locus can be determined without any transfor- 
mation of co-ordinates whatever. 
For from the relation 

4AB-C 2 = (20) 

it follows that, in case the condition 

CE-2BD = (21) 

is satisfied, then the condition 

CB-2AE=0 (22) 

must also be satisfied (O being supposed not equal to zero). 
That is, the numerators of h and h must vanish together ; 
whence it also follows from equation (3) that in this case the 
condition A = 



212 ANALYTIC GEOMETRY. 

is satisfied. Therefore the locus must consist of straight 
lines. 

The equations of these lines are given immediately by equa- 
tion (2), which may now be written 



2By+Cx + E± V£" 2 - 4 BF= 0. (23) 

We see that the locus consists of two parallel straight 
lines, which are real, imaginary, or coincident according as 
E 2 — 4 BF is positive, negative, or zero. 

This result may also be obtained by solving the general 
equation with respect to x, and then introducing the condi- 
tions of (20), (21), and (22). We thereby obtain the equation 

2Ax+Cy + D± ->fi7 — ±AF= 0. (24) 

This equation, by means of (20)-(22), may be shown to 
be identical with (23). 

When, in addition to the conditions given in (20)-(22), 
E 2 = ±BF, it follows that D 2 = 4:AF, and the locus is the 
single straight line represented by either of the equations 

2Ax+Cy + B = ) 
2By+Cx+E = 0. 

If the numerators of h and k are not zero, the centre is at 

infinity. Although we cannot now transform the origin to 

the centre, we can make the term involving xy disappear by 

proceeding exactly as in § 201 ; that is, by turning the axes 

through an angle 6, the value of which is determined by the 

equation n 

1 tan 20 = — - — (25) 

A-B v ; 

If P, Q, U, V represent the new coefficients of x 2 , y 2 , x, y, 
respectively, P and Q will have values identical with those 
of P and Q given in § 201, and 

U= DcozO+Fsine, (2G) 

V= -DsinO + F cos 6. (27) 



LOCI OF THE SECOND ORDER. 213 

The relations of P, Q, A, B, and C, found in § 201, also 
hold true, namely, 

P+Q=A+B, 
P-B=± -V(A - £) ! + C\ 

where the sign before the radical should be the same as that 
of C. But since now C 2 = 4 AB, the value of P— Q becomes 

P-Q = ±(A+B); 

whence, if C is negative, 

P=0, Q=A+£. 

But if C is positive, 

P = A+B, Q = 0. 

Suppose that C is negative. Then the general equation 

becomes Qy* + Ux+ y y+F= . (28) 

Divide by Q, and we have 

V A — xA — v A — =0, 

*,U,V 2 ,Uf,F V 2 \ n 
y+ Q y + ^ + Q{ X + U-lQUr Q ' 

If we now take as a new origin the point 

4 QF-V 2 __T\ 
4c7Q ' 2QJ' 

equation (28) becomes 

U 
U Q 

which represents a parabola whose axis coincides with the 
axis of x, and is situated on the positive or the negative side of 



214 ANALYTIC GEOMETRY. 

the new origin, according as U and Q are unlike or like in 
sign (§ 103). 

The vertex of the parabola is the new origin, and the 
parameter is equal to the coefficient of x in the equation of 
the curve. 

Suppose that C is positive. Then the general equation 

becomes p x 2 ±Ux + Vy + F= 0. (29) 

And this, by changing the origin to the point 

4FP ' 2P 
becomes x 2 = — -y. 

This represents a parabola having the axis of y for its axis, 
and placed on the positive or the negative side of the new 
origin, according as T^and P are unlike or like in sign. 

We have already found that the value of P or Q, when not 
zero, is A + B. 

We may obtain the values of U and V in terms of the 
original coefficients, as follows : 

From (25) we find, by Trigonometry, 

Introducing the condition AAP = C 2 , we obtain 

2A 

tan 6 = — ^—i if C is negative ; 

= > if C is positive ; 

o 

whence, if Cis negative, 

sin 6 = — - i cos0 = — - • 

V4^ 2 +(7 2 V4^4 2 + C 2 



LOCI OF THE SECOND ORDER. 215 



And if C is positive, 

$mv = — - > cos#=- 



■y/4:B 2 +C 2 V4,8 2 + C 2 

By substitution we obtain from (26) and (27) 

u= 2 As-cn (80) 

■V4tA 2 +C* 
r= C U-2BD i (31) 

V4^ 2 +C 2 
the positive sign being placed before both radicals. 

204. Special Cases. When certain of the coefficients in 
equation (1) are equal to zero, the preceding investigations 
may require some modification as to details, not as to results. 

Only three special cases require any notice. 

1. Suppose that B = 0. In this case we cannot solve the 
general equation with respect to y, and hence find the value 
of A, as has been done in § 199. If, how r ever, we solve the 
equation with respect to x, we obtain 

x =- gM £r ± iz^ L ' xi + M ' x + JV '< (32) 

where 
L' = C 2 -4:AB, ir = 2{CD-2AE) ) JY 1 =B' 2 - ±AF; 

and by proceeding as in § 199, w T e obtain exactly the same 
value of A as before. 

In this case 2 cannot be zero ; so that the locus always 
belongs to the first class of curves. 

2. Suppose that ^4 = i? = 0. The general equation now 
becomes Qxy + Dx+Ey+F= . (33) 

In this case the value of A cannot be found directly by the 
method of § 199. If, however, (33) represents straight lines, 



216 ANALYTIC GEOMETRY. 

the equation may be written as the product of two linear 
factors > (Px + M) (Qy + N) = 0. (34) 

Equating coefficients in (33) and (34), we obtain - 

FQ = C, PJST=D, QM=E, MN=F, 
whence CF= DE. 

When this condition is satisfied, (33) represents two straight 
lines, and their equations are 

Cx+E = 0, 
Cy+D = 0. 

The two lines are parallel to the axes, and therefore per- 
pendicular to each other. 

These results may also be obtained by putting A = and 
B == in the results reached by the general investigation. 

In general, since 2 is negative, equation (33) represents an 
hyperbola. By changing the origin to the centre, the equa- 
tion takes the form ^ = aconstant) 

which we know (§ 194) represents an hyperbola, referred to 
its asymptotes as axes. Therefore, in general, equation (33) 
represents an equilateral hyperbola. 

3. Suppose 2 = 0, and also C= 0. Then either A or B 
must also be zero. A and B cannot both be zero, for in this 
case equation (1) would cease to be an equation of the second 
degree. 

If J. = 0, equation (1) becomes 

By*+I)x + Ey+F=0, 

an equation of the same form as (28). Therefore the locus is 
a parabola. 

If B=0, equation (1) becomes 

Ax 2 +Dx + Ey + F= 0, 

which has the same form as (29), and the locus is a parabola, 
with its axis on the axis of y. 









LOCI OF THE SECOMD ORDER. 



217 



205. The main results of the investigation are given in the 



following Table : 



loci represented by the general equation of the 

Second Degree, 

Ax 2 + By 2 + Cxy + Dx + Ey + F= 0. 


CLASS. 


CONDITIONS. 


LOCUS. 


I. 

Loci 

having a 

centre. 


2 positive, A not zero. 
2 positive, A = 0. 
2 negative, A not zero. 
2 negative, A = 0. 


Ellipse (real or imaginary). 

Point. 

Hyperbola. 

Two intersecting straight lines. 


II. 
Loci not 
having a 

centre. 


2 = 0, A not zero. 
2 = 0, A = 0. 


Parabola. 

Two parallel straight lines. 



Thus it appears that there are no loci of the second order 
besides those whose properties have been studied in the pre- 
ceding chapters. 

206. Examples. 1. Determine the nature of the locus 
represented by the equation 

7x 2 -Ylxy + 6y 2 + <23>x- 2y + 20 = 0. 

Here we have 2 = — 121, A = 0. 

Therefore the equation represents two intersecting straight 
lines. By substitution in the values of h and k (§ 200), we 
find that the lines intersect at the point (2, 3). 

If we change the origin to the centre, the equation of the 

lines become 

7^-17a;y + 6?/ 2 = 0. 



218 ANALYTIC GEOMETRY. 

2. Determine the nature of the locus represented by the 
equation *> x 2 + by 2 + 2xy -I2x -\2y = 0, 

and reduce the equation to its simplest form. 

Here 2=96, A = 864, h = 1, i = l, 

j?=12, P=6, Q = 4. 

Therefore the locus is a real ellipse, the centre is the point 
(1, 1), and the equation, in its simplest form; is 

Sx 2 + 2y 2 = 6. 

The value of 0, found from equation (10), is 45°. 
Therefore the equations of the new axes of x and y referred 
to the original axes of co-ordinates, are respectively 

x — y = 0, 

x + y - 2 = 0. 

The form of the reduced equation shows that the major 
axis of the ellipse is situated on the axis of y. 

3. Determine the nature of the locus of t\iQ equation 

x 2 + y 2 - bxy + 8x— 20y + 15 = 0, 
and reduce the equation to its simplest form. 
In this case 

S--21, A =-21, h =-4, A = 0, 
5 = 1, P=-f Q = i, = 45°. 

Therefore the locus is an hyperbola ; and since H agrees in 
sign with Q, the transverse axis is situated on the new axis 
of y. 

The equation of the curve, in its simplest form, is 

7 y 2 -3* 2 = 2. 

And the equations of the axes of the curve referred to the 
original axes of co-ordinates are 

x — y + 4 = 0, 

* + y + 4 = 0. 



LOCI OF THE SECOND ORDER. 219 

4. Determine the nature of the locus of the equation 

x 2 + y 2 - 2x7/ + 2x — y - 1 = 0, 

and reduce the equation to its simplest form. 

Here 2 = 0, A = l, P=0, Q = 2, = 45°. 

Therefore the locus is a parabola, the axis of which coin- 
cides with the new axis of x. 

V2 

From equation (30) we have U= 

A 

Hence the equation of the curve, in its simplest form, is 

Since £7 and Q agree in sign, the parabola is situated on 
the negative side of the new origin. 

After the original axes of co-ordinates have been turned 
through the angle ^ _ . r 

the vertex of the parabola is the point 

/^ 3V2\ 

V8V2 8 / 
and the equation of its axis is 

3V2 

If the axes of co-ordinates are turned through the angle 

6 = - 45° 

to their original position, the vertex becomes the point 

/19 31\ 
\T&' 16/ 

and the equation of the axis becomes 

Ax- 4y + 3 = 0. 



220 ANALYTIC GEOMETRY. 

207. The locus of a point which so moves that its distance 
from a fixed point bears a constant ratio to its distance from a 
fixed straight line, is called a Conic. 

The fixed point is called the Focus ; the fixed straight line, 
the Directrix ; the constant ratio, the Eccentricity. 

If we denote the distance from the focus to the directrix by 
d, the eccentricity by e, and take for the axis of x the perpen- 
dicular from the focus to the directrix, and for the axis of y 
the directrix, the equation of a conic is easily found to be 

y 1 -f- (% — d) - = e 2 x 2 , 

or (1 - e 2 )x 2 + y 2 -2xd+d 2 = 0. 

This is an equation of the second degree ; hence a conic is 
always a locus of the second order. 

If e — 0, the curve is a circle (§ 70). 
If e < 1, the curve is an ellipse (§ 162). 
If e—1, the conic is a parabola (§ 101). 
If e > 1, the conic is an hyperbola (§ 195). 

In many treatises the properties of these curves are deduced 
from the definition above given. 

208. The term " conic " is an abbreviated form of " conic 
section." The four curves, circle, parabola, ellipse, and hy- 
perbola, were originally called conic sections, because it was 
discovered that they could all be obtained by making a plane 
cut the surface of a cone of revolution in different wa}^s. 

If the plane be perpendicular to the axis of the cone, the 
section made by the plane will be a circle. 

If the plane be parallel to an element of the surface, the 
section will be & parabola. 

If the plane make a greater angle with the axis than the 
elements make, the section will be an ellipse. 

If the plane make a less angle with the axis than the ele- 
ments make, the section will be an hyperbola. 



LOCI OF THE SECOND ORDER. 221 

Still further : 

If the plane pass through the axis, the section will consist 
of two intersecting straight lines. 

If the plane pass through the vertex perpendicular to the 
axis, the section will be & point. 

If the plane be parallel to an element, and we conceive the 
vertex removed to an infinite distance, the cone will become 
a cylinder, and the section will consist of two parallel lines. 

Hence conic sections, in the most general sense of the term, 
embrace all loci of the second order. 

Ex. 40. 

Determine the nature of the following loci, and reduce each 
equation to its simplest form. 

1. 3x 2 + 2y 2 — 2x + y-l=0. 

2. l+2x+Sy 2 = 0. 

3. y 2 -2xy-\-x 2 -Sx + 16 = 0. 

4. 3x 2 + 2xy+Sy 2 -16y + 23 = 0. 

5. x 2 -10xy + y 2 + x + y + l = 0. 

6. o^-2xy + y 2 -6x-e>y + 9 = 0. 

7. x 2 + xy+y 2 + x + y-b = 0. 

8. y 2 -x 2 -y=0. 

9. 36^ 2 + 24^/ + 29y 2 -72^ + 126y + 81=0. 

10. y 2 -2^-8y + 10 = 0. 

11. 4^ 2 + 9y 2 + 8.r + 36y + 4-0. 

12. 52^ 2 + 72^ + 73y 2 = 0. 

13. 93/ 2 -4^ 2 -8.r + 18y + 41 = 0. 

14. y 2 — xy — 5 x + by = 0. 

15. 16^ 2 -24a:y + 9y 2 -75^-100?/ = 0. 

16. 4^ 2 + 4^y + y 2 + 8^ + 4y-5 = 0. 



ANSWERS. 



Ex. 3. Page 7. 

1. Let x t = — 2, y x = 5, a? 2 = - 8, y 2 = - 3. Substituting in [1], we 
have 

d = V(- 6) 2 + (- 8) 2 = VlOO = 10. 

In Fig. 3 the points P and Q are plotted to represent this case. If 
we choose to solve the question without the aid of [1], we may neglect 
algebraic signs, and we have 

QR = NO - M = 8 - 2 = 6 ; 
PR = PM + MR = 5 + 3 = 8 ; 
... PQ2 = Qjp + p E 2 = 36 + 64 = 100, and PQ = 10. 

2. 13. 8. 5, 5, 6. 

3. 5. 9. a, 5, V(?TF. 

4. 10. 10. V29, 5, 2VT0, 4V5 ; 

5. 2V^+T 2 . 2VT0, 9V2. 

6. 25, 29, 20 V2. 11. 8 or -16. 

7. 2VT7, 5V2, V106. 12. (x - If + (y - 2) 2 = 121. 

13. (x - 7) 2 + (y - 3) 2 = (x - 4) 2 + (y - 5) 2 , which reduces to x + y = 7. 

Ex. 4. Page 9. 

1. (6,6). 3. (2,-2). 5. (7, 1). 

2. (-1, 0). 4. (3, -1), (J, Y), (-J, -f). 6. (a, -5). 
7. Take the origin of co-ordinates at the intersection of the two legs, 

and the axes of x and y in the directions of the legs. Then, if a and b 
denote the lengths of the legs, the co-ordinates of the three vertices will 
be (0, 0), (a, 0), and (0, b). 

10. Observe that now the distances RB and BQ will be x — x 2 and 
V-Vv 12. (f,i). 14. (71,-311). 

11. (6, 2). 13. (4, 8). 15. (13,-1). 



ANALYTIC GEOMETH' 









Ex. 7. 


Page 23. 


1. 


12, 16. 






14. 


Locus does not cut the axes. 


2. 


-10, 6. 






15. 


(5, 7). 


3. 


± 4, ± 4. 






16. 


(2, 1). 


4. 


+ f,±2. 






17. 


(3, -4) and (-4, 3). 


5. 


± f , imaginary. 




18. 


(3, 4). 


6. 


±f,-4. 






19. 


(5, 3) and (3, 5). 


7. 


± 5, ± a. 






20. 


(0, 0) and (2, 4). 


8. 


3 on OX 






21. 


(5, -3), (6, 4), (-1,-4). 


9. 


± 3 on OX 






22. 


V61, V265, Vl04. 


10. 


Locus passes 


through 


origin. 


23. 


3, 4, 5. 


11. 


Locus passes 


through 


origin. 


24. 


(a, 5) (— a, b), (—a, — 5) (a, - 


12. 


( On OX, 8, 
I On OF, 2 


and - 4. 

tVis. 




25. 
26. 


No. 
10. 


13. 


Locus passes 


through 


origin. 







•b). 



Ex. 9. Page 31. 

1. Let x and y denote the variable co-ordinates of the moving point. 
Then it is evident that for all positions of the point # = 3y. There- 
fore the required equation is x = 3y or x — Sy = 0. Does the locus of 
this equation pass through the origin ? 

2. a- 6 = 0, x + 6 = 0, x = 0. 

3. y — 4 = 0, y + l = 0, y = 0. 

4. The line z = 3 is the line AB (Fig. 74); how is this line drawn? 
The locus of the variable point consists of the two parallels to AB, drawn 
at the distance 2 from AB. Let CD, EF, be these parallels, and (x, y) 
denote in general the variable point, then for all points in CD x = S 
+ 2 = 5, and for all points in EF x = 3 — 2 = 1. Therefore the equation 
of the line CD is a<~5 = 0, and that of the line EF is z-l = 0. The 
product of these two equations is the equation (x — b) (x — 1) = 0. This 
equation is evidently satisfied by every point in each of the lines CD 
and EF, and by no other points. Therefore the required equation 
is (x — 5) (x — 1) = 0, or x 2 — 6 x + 5 = 0. Verify that this equation is 
satisfied by points taken at random in the lines CD and EF. 



ANSWERS. 6 

5. y 2 — 11 y + 24 = 0, two parallel lines. 

6. z 2 -f-8z-9 = 0, two parallel lines. 7. £ + 3 = 0, y-2 = 0. 
8. It is proved in elementary geometry that all points equidistant 

from two given points lie in the perpendicular erected at the middle 
point of the line joining the two given points. This perpendicular is 
the locus required, and its equation evidently is x = 3. 



Y 




F 


B 


D 







E 


A 


X 
C 



Y 


/ 


P 







A X 



Fig. 74. 



Fig. 75. 



Let us now solve this problem by the analytic method. Let 
(Fig. 75) be the origin, A the point (6, 0), and let P represent any posi- 
tion of a point equidistant from and A, x and y its two co-ordinates. 
Then from the given condition 

PO = PA. 
Therefore x 2 + y 2 = (x - 6) 2 + (y - 0) 2 , 

or x 2 + y 2 = x 2 -I2x + 36 + y 2 ; 

whence x = 3, 

the equation of the locus required. 

9. ^-1 = 0. 10. y-2 = 0. 11. x -y-l = 0. 12. x-y = 0. 

13. x 2 + y 2 = 100, a circle with the origin for centre and 10 for radius. 

14. Express by an equation the fact that the distance from the point 
(x, y) to the point (4, —3) is equal to 5. The equation is (x — 4) 2 
+ (3/ + 3) 2 = 25. 

15. (x + 4) 2 + (y + 7) 2 = 64. 16. x 2 + y 2 = Sl. 

17. Draw AO ± to BC (Fig. 76). Take AO for the axis of x, and 
BC for the axis of y ; then A is the point (3, 0). 

Let P represent any position of the vessel, x and y its co-ordinates 



4 ANALYTIC GEOMETRY. 

OM and PM. Join PA, and draw PQ JL i?C, and meeting it in Q. 
Then from the given condition 

PA = PQ=OM. 
Therefore PA 2 = OM 2 . 

Now PA 2 = AM 2 + PM 2 = (x - 3) 2 + y 2 , and OM 2 = a: 2 . Substituting, 

wehave (s-3)' + 3 ,W; 

whence y 2 = 6 a; — 9. 



B 


i> 


Q 







C 


I A M 



Fig. 76. 



The locus is the curve called the parabola. We leave the discussion 
of the equation as an exercise for the learner. 

18. If BC is taken for the axis of y, and the perpendicular from A to 
BC as the axis of #, the required equation is y 2 = \2x — 36. 

19. x 2 — 3y 2 = 0, two straight lines. 

20. x 2 +y 2 = K 2 - a 2 , a circle. 

21. 4:ax ± K 2 = 0, two straight lines. 

Ex. 10. Page 33. 

4. d = Va?! 2 + y^ 2 . 6. a; + y = 7. 

5. (& - 4) 2 + (y - 6) = 64. 7. (- 1 /, f) ; I V2. 

8. Take two sides of the rectangle for the axes, and let a and b rep- 
resent their lengths ; then the vertices of the rectangle will be the points 
(0, 0), (a, 0), (a, b), (a, b). 

9. Take one vertex as the origin, and one side, a, as the axis of x ; 
then (0, 0) and (a, 0) will be two vertices. Let (b, c) be a third vertex ; 
then (a 4 b, c) will represent the fourth. 



ANSWERS. 



10. 


(11, 2). 11. 


(5,-2),(|,-i),(|,-V). 


12. 


(1,-|). 13. VT7. 14. 


(i, J). 16. (6, 23). 


17. 


(x l + 3x 1 y x + 3y 2 \ ^1 + 2*2 
{ i 4 / \, 2 


2/1 + 2/2^ /3^ + ^ 2 3y x + y 2 
2 / \ 4 ' 4 


21. 


3 or -23. 23. 


(4, 8) and (4, - 8). 


22. 


1 3 and 2 on OX. 24. 


(2 a, a) and (—2 a, a). 



1 6 and 1 on Y. 25. (a, 0) and (- a, 0). 

26. 10, Vl04, V52. 

27. Taking the fixed lines for axes, the equation is y = 6a? or x = 6y. 

28. Taking A for origin, and JL^ for the axis of x, the equation is 
x - VZy = 0. 

29. Taking the fixed line and the perpendicular to it from the fixed 
point as the axes of x and y respectively, the required equation is 
a; 2 -4y 2 = (y-a) 2 . 

Ex. 11. Page 40. 

1. z-y+l = 0. 20. y+3 = 0. 

2. 2a? -y- 3 = 0. 21. x- 2 = 0. 

3. x + y -1 = 0. 22. x -y +2 = 0. 

4. z-y = 0. 23. z-y + 5 = 0. 

5. 3a; + 2y-12 = 0. 24. a?-t/-4 = 0. 

6. 2*-3y + 6 = 0. 25. a: - V3y - 4V3 = 0. 

7. 6z-y-7 = 0. 26. y + 4 = 0. 

8. 4z-3y = 0. 27. V3z-y-4 = 0. 

9. y = 0. 28. 2 = 0. 

10. y = 4. 29. V3;r + y + 4 = 0. 

11. 5z-2y=0. 30. z+y + 4 = 0. 

12. nx-my = 0. 31. 2 + V3y + 4>/3 =0. 

13. a;-y-3 = 0. 32. z-y-4 = 0. 

14. V3x-y +7-2>/3 = 0. 33. 3z + 4y-12 = 0. 

15. a:- y +14 = 0. 34. a;-3y-6 = 0. 

16. V3z + 3y + 12-13 V3 = 0. 35. x + y + 3 = 0. 

17. x-v / 3y-4 = 0. 36. 3a; -5y -15=0. 

18. a + y-3 = 0. 37. aj-2y+10 = 0. 

19. V3z + y=0. 38. or -y -1=0. 



45. 



46. 



6 ANALYTIC GEOMETRY. 

39. x _y_ n = o t 52. x + y + 6V2 = 0. 

40. 4a: + y-4n = 0. 53. (1,7). 

41. a;-'ry-5\/2 = 0. 54. (1, 2). 

42. x - y V3 + 10 = 0. 55. (2, 1). 

43. a:+yV3+10 = 0. 56. (3, 2). 

44. x-yVS-10 = 0. 57. (-2,-6). 
cx+7y + ll = 0, x-3y+l=0, 58. (-4,6). 
j 3^ + 2/-7 = 0. 59. (5,-3), (6, 4), (-4,-1). 
ra:-72/=39, 9a-5y = 3, 60. 9* + 2y = 0, |V85. 
1 4a; + 3/ = 11. 61. y — % = y L — x v 

cl7x — 3y = 2o, 7x + 9y c (d—c)x -(b — a)y = ad—bc, 

1 =-17, 5rc-6y-21 = 0. ' \ (d-c)x + (b-a)y = bd-ac. 
(bx — 3/ = 0, 5^ + 63/ — 35 = 0, f 2 y 2 £ + (^ — 2 # 2 ) y — x Y y 2 = 0, 

48. -j 3 x — y = 21, 9 x + 4 3/ = 0, 64. -j ^as + (2 x 1 — x 2 ) y — x x y 2 = 0, 
[y = 0, 14a; + 3 t y = 29. { y 2 x — (x 1 + x 2 ) y = 0. 

49. z-7/\/3-7i = 0. 65. m = 4. 

50. 3/ = .t + 3. 66. m = 3. 

51. z + 2/-6\/2 = 0. 67. b = -9. 

68. 3/3 ~ 3/i = y 2 ~ 2/i ? 0Ij ^(3/2 -y 3 ) + ^(2/3-3/!) + ^3(3/! -2/ 2 ) = 0. 



Ex. 12. Page 45. 

1. a = 4, 6 = 7, m = — J. 12. a = & = 0, ra = J. 

2. a = 27, 5 = - 9, m = J. 13. a = 6 = 0, m = 3. 

3. a = -f, & = 1, ra = f. 14. a = - 4, 5 = 5, m=f. 

4. a = Z> = 0, ra = f. 15. a = -2, 6=12, ?n=6. 

5. a=5 = 0, m = -{. 16. a = 6, 6 = - 6, m = l. 

6. a = 8, Z>=-6, m = f. 17. a=-10, b^-^ -Vs, m=— jV3. 

7. a = Z> = 3, m = -l. 18. a = 10, 6 = ^V3, m = |V3. 

8. a— 11, 6 = — y,m=-l 5 

20. m = . 

9. a = — 3, b = 5, ra = J. a 

10. a =2, & = -3, m = §. 21. ; - 8. 

11. a=-2, 6 = -3, m = -|. 22. ^4 = - 4, £ = - 1. 



ANSWERS. 
24. \ A= ^ 2 " y^ B== ~ ^ 2 ~ x ^ 25. m = y2 "'- Vl 6 = ^3/i ~ *i3/ 2 > 

I C = tf^ — iTj^j* x 2 ~ X l X 2 ~ X l 

27. (a) cos a = f|, sin a = ^, p = 2. 
(5) cos a = — |f, sin a = — T \, p = 2. 

(c) cos a = Jf , sin a = — T 5 3, p = 2. 

(d) cos a = — |f, sin a = T 5 3, p = 2. 



Ex. 13. Page 49. 

1. 3 a? -y- 1.6 = 0. 6. a + 4y + 49 = 0. 

2. 3;z-4y-3=0. 7. 7z - 23y +193 = 0. 

3. 4z-y = 0. 8. y = 2x. 

4. y — 8 = 0. 9. 7a? + 5y-ll=0. 

5. x _5 = o. 10. cc + 33/ — 5 = 0. 

Ex. 14. Page 51. 

2. tan <b = J. 3. tan <f> = A. 4. tan <p = — - — . 

n 2 + 2 
5. 90°. 6. 45°. 7. 90°. 8. 0°. 9. 30°. 



n jy = 5z-10, 13 jy-3 = m'(.r-2), 

U + 5y = 28. I. andm' = -(5±3V3). 

12 {y = 5# + ll, /y — 3 = ra / (z — 1), 

' l.+sy-s-a 14. j andm/a H ± eV3. 

22. 2z + 3y-31 = 0. 



11 



23. 62z + 31y -1115 = 0. f *~3y + 26 = 0, 

30. J 5z + 3y + 8 = 0, 



24. y = 6z-27. [ o _i_ * Q n 
■* v 2 a; + 3 y — 9 = 0. 

25. y = ma; ± c? Vl + m 2 . 01 ~ A 
^ 31. x — 6 = 0. 

26. 2fc = ^,-5). ^_ %+ 12 = a 

27. ax-by=a 2 -b\ 32 . J 10*- 4y + 63 = 0. 

28. (a±%-(a + &)(z-a) = 0. I lSx-^0y + 111 = 0. 

( x— y — 6 = 0^ meeting in the point 
33. J 2x- y- 2 = t (-4, -10). 
[ 5a;_3y_ 10 = J Distance = V85. 

ok - -4 ± B tan ^ , „ n 
35. y — y x = — r (x — a^). 



ANALYTIC GEOMETRY. 

Ex. 15. Page 57. 
|V5. 3. 4. 4. ^VlS. 5. 0. 

The learner should construct the given line, 
and observe how the sign of the required dis- 
tance changes when we cross the line. 

8. 6, 5, 4, 3, 2, 1, 0, — 1. The learner should construct the lines, and 
observe the change of sign of the distance, as in No. 7. A study of 
Nos. 7 and 8 will make it evident that, in equation [12], if Ax 1 -\-By l + C 
has the same sign as C, then the point (x v y x ) lies on the same side of the 
line as the origin, and vice versa. 



1. 


H 


'10. 




2. 


7. 


24 

5 ) 


20 
~5~» 


_1_6 
5 ' 


12 

5 ' 




8 
5' 


4 
5> 


0, 


4 
~ 5' 




JL5 
5 ' 


JL2 

5 » 


9 
5' 


1, 




3 

5' 


0, 


_ 3 
5' 





9. 


fVio. 


10. 


f?V4l. 


11. 


2\/2. 


12. 


*VT8. 


13. 


i i 


14. 


ftVIE. 


15. 


^V2. 


16. 


iV2. 



17. 
18. 


Va 2 + b\ 

ab Sab 


22. 
23. 

24. 
25. 

26. 


4. 


19. 


\/a' + b' 2 Va 2 + 5 2 
a 2 -b 2 


VA 2 + B 2 

fVio. 


20. 


Va 2 + b 2 
Ah+BJc + C-D 


AVFb. 




VA 1 + B 2 
2. 


Sab 


21. 


2\/d 2 + 6 s 



Ex. 16. Page 60. 



1. 


1J. 4. 40. 


8. 


35. 




11. 


26. 


2. 


12. 5. ab. 


9. 


19J. 




12. 


96. 


3. 


29. 7. 26. 


10. 


i(*i2/ 2 - 


- X 2V\)- 


13. 


61. 


14. 
15. 
16. 


i(a-e)(b-l). 
\(a-b){2c-a-b). 


21. 
22. 


9 a 2 . 

2 c 2 
21* 




27. 
28. 


2 m 
hab. 


17. 

18. 


60° > 60°,60°;9V3. 
10. 


23. 
24. 


24. 

36. 




29. 


C 2 
2AB 


19. 


i 

4- 


25. 


16. 




30. 


56. 


20. 


1J. 


26. 


iab. 




31. 


10|. 



ANSWERS. 



Ex. 17. Page 62. 

3. 2, oo, 90°, 2, 0°. 4. 0, 0, 45°, 0, 135°. 

5. IV3-2, 2V3-1, 60°, 2 I" 1 ' ISO - 

6. 2, |V3, 150°, 1, 60°. 26. 4y = a; + 8. 

7. 2, -|V3, 30°, 1, -60°. 27. 4y = 9z-24. 

8. W3, -2, 60°, 1, -30°. 28 I 9x ~ 20y + 96 = °' 

' ' n l5o;- 4^ + 32 = 0. 

Q , + 23 y- 18 = 0, ^ 

9 ' I 49* + 7y-82 = 0. 29. 88a. -121 y + 371 = 0. 

10 AV82 30. \ 5x ~ 3/"10 = 0, 



11. 



3z + 4y-57 = 0, 

12z-5y-39 = 0, l x " V U 

12,-5y + 24 = 0. 32 - x±y-5 = 0. 



20. 



area = 63. 33. 2z = y, 2y = x. 

12. 43. 34. 4, + 5y + 11±3V41=0. 

13. z = 3. 35. y = (l±V2)(, + 2)._ 
i4 f a ._ y + l = > 36< 13y~12 == 23±5V29 



s + y-l = 0. 13,-30 14 

15. 5, + 6y-39 = 0. 37 f7,-3y + 15 = 0, 

16.14,-33,-30 = 0. •l3, + 7 3 /-19 = 0. 

« Q J 8, + 7y-19 = 0, 

17. 4,-5y + 8 = 0. 38. ( 16jB + 3 J + 17 _ a 

18. a; + 3/ -7 = 0. 39, 450^ 

19. y-ys _ V2-yi _ 40. 90 . 

a;-o? 3 z 2 -s, ^ 31y/26 

y = 3, 13y = 5a;-l 1 ' 143 

9 2/ = 5 ^ + 7. 42 6^ + gjfe-i 

21. 92, + 69y + 102 = 0. ' y/ *~+b 2 ' 

22. , + 4y = 34. 43 c 2 

23. 3, + 4y-5a = 0. V/^+T 2 

24. 3, + 4y = 24. ^ a ,- 

J 44. — Vl + ra 2 . 

y m 

25. v-ih—'-Mx—b 45 _ c2> 



10 ANALYTIC GEOMETRY. 



46. 


k 2 

e' 

2a 2 + 5aZ> + 2Z> 2 


54. 


xy represents the two axes. 


47 


57. 


a = 5. 


48. 
49. 
50. 


6 
17*. 

59. 


58. 
59. 
60. 


a? + a = 0, # — & = 0. 
a? + a = 0, 2/ + 6 = 0. 
The axes and x = y. 


51. 


(10, 5J). 


61. 


2x — y = 0, 7tf + y = 0. 



62. If /i denotes the altitude of the triangle, and the base is taken as 
the axis of x, the locus is the straight line y = h. 

63. The equation of the locus is 

(x - x x f + (y - y,f = (x- xj> + (y - y,f. 
This is the equation of the straight line bisecting the line joining (x v y^ 
and (a? 2 , y 2 ), and J_ to it. 

64. The two parallel lines represented by 

Ax+£y + C± dVAF+7? = 0. 

CK , 7 c ~ Ax + By + C . A?xiJ¥y + & t 

65. a; + v = k. 66. ^ + * = a;. 

v^i 2 +£ 2 \/A /2 +B' 2 

67. Let b denote the base, k 2 the constant difference of the squares of 
the other two sides. Taking the base as axis of x, and middle point 
of the base as origin, the equation of the locus is 2 bx = k 2 . 

Ex. 18. Page 70. 
1. 7x + y = 0. A fff._y-i.8-0, 6 - 64a;-23y = 59. 



( x -M 



2. z + 2?/ + i3 = 0. l_+2/-6_=0. 7. 44z + </ = 0. 

3. 5_ + 6y-37=0. 5. y-=a + 3. 8. 5z + 2/-16 = 0. 

9. (AW -A'C)x + {B& -B'C)y = 0. 

10. (.B^-il-JOy+C- 17 -- 1 ^^ - 
u Ar + _??/ + £ _. A'x+B'y + C? 

' Ax^ + By^C A'x x + B' yi + C' 
12. 472z -29y + 174 = 0. 13. y =_V3 + 3 - V3. 

14 f 4a; + 32/ — 25 = 0, y „_ m&-g 

(3^-4y + 25 = 0. 'a 5 ma + b 

16-18. Generally the easiest way to solve such exercises as these is 
to find the intersection of two of the lines, and then substitute its co- 
ordinates in the equation of the third line. 



ANSWERS. 11 

in i on rri m // — m b"—b 

19. m = l. 20. when = 

m //_ m / &// _&/ 

21. If we choose as axes one side of the triangle and the correspond- 
ing altitude, we may represent the three vertices by (a, 0), (— c, 0), (0, b). 

22. Choosing as axes one side and the perpendicular erected at its 
middle point, the vertices may be represented by (a, 0), (— a, 0), (o, c). 

23. It is well here to choose the same axes as in No. 21. 

24. Choosing the origin anywhere within the triangle, it is evident 
that the equations of the bisectors in the normal form may be written as 
follows : 

(x cos a -\- y sin a — p) — (x cos a/ + y sin a/ — p') = 0, 

(x cos a' + y sin c/ —p f ) — (x cos a 7/ + y sin a /7 —p // ) = 0, 
(x cos a" + y sin a // —p // ) — (x cos a + y sin a — _p) =0. 
Now, by adding any two of these equations, we obtain the third; 
therefore the three bisectors must pass through one point. 

25. ( 2V 2. VT0,2V10. ^ f s + y _ 2 = 0, 

I Origin within the A. { x _ , 



-y-14 = 0. 



26. -yyiO, !?-V34, i|Vl3. 

on X - 1 = 0, 

27. f«+y+w-o, 30. I 

7z-7y + 23 = 0. * 



28. 



7z-9y + 34 = 0, 31. y-ms- & = ± ,y--m'a? -&' 

9z+7y-12 = 0. Vl + m 2 V1 + ?ti/ 2 



Ex. 19. Page 74. 

1. (i.) Parallel to the axis of x, (ii.) parallel to the axis of y. 

2. When ad = be. 

3. The two lines are real, imaginary, or coincident, according as 
H 2 — 4 AB is positive, negative, or zero. The two lines are JL to each 
other when A + i? = 0. 

5. a? 4- y + 1 = 0, and x — Sy + 1 = 0. 

6. x-2y ±(y-3)V^l = 0. 

7. z-y-3 = 0, and *-3y + 3 = 0. 8. 45°. 9. JT= 2. 

11. iT=28. 12. K=-\ 5 -. 



12 ANALYTIC GEOMETRY. 

Ex. 20. Page 76. 

1. Take the point as origin, and the axis of y parallel to the given 
lines. If the equations of the given lines are x = a, x = b, and if the 
slopes of the lines drawn in the two fixed directions are denoted by 
m / , m // , the equation of the locus is 

(b — a)y = m'b (x — a) — m // a (x — b). 

2. If a and b are the segments of the hypotenuse made by a perpen- 
dicular dropped from the vertex of the right angle, the equation of the 

locusis ¥-■■*/"■ 

3. Let OA = a, OB = b. Then the equation of the locus is x + y = a + b. 

4. Take as axes the base and the altitude of the triangle. Let b 
denote the base, a one segment of the base, h the altitude. Then the 
equation of the locus is 2 x 2 y 

b — a h 
This is a straight line joining the middle points of the base and the 
altitude. 

5. Take as axes the sides of the rectangle, and let a, b denote their 
lengths. The equation of the locus is 

bx — ay = 0. 
Hence the locus is a diagonal of the rectangle. 





Ex. 21. 


Page 79. 


1 


x 2 + y 2 = — 2 rx. 


13. 


(4, 0), 4. 


2. 


x 2 + y 2 = 2 ry. 


14. 


(-4,0), 4. 


3. 


x 2 + y 2 = — 2 ry. 


15. 


(0, 4), 4. 


4. 


(re - 5) 2 + (2/ + 3) 2 = 100. 


16. 


(0,-4), 4. 


5. 


x 2 + {y + 2) 2 = 121. 


17. 


(0, I), f. 


6. 


(x - 5) 2 + y 2 = 25. 


18. 


(0, 0), 3 k 


7. 


(x + 5) 2 + y 2 = 25. 

(x - 2) 2 + {y- 3) 2 = 25. 


19. 
20. 


(0, 0), 21. 


8. 


(0, 0), Va 2 + b 2 . 


9. 
1 1 


x 2 + y 2 -2hx~2ky = 0. 
(1, 2), V5. 
(1, 1), iV38. 


21. 


(| °}|V5. 


11. 
12. 


22. 


/fc k\ Vh 2 + k 2 



ANSWERS. 



13 



23. When D = D' and E=E f \ in other words, when the two equa- 
tions differ only in their constant terms. 

24. In this case r = 0. Hence the equation represents simply the 
point (a, b). We may also say that it is the equation of an infinitely 
small circle, having this point for centre. 

. (h 7\ 5J9. 



26. 



27. 



28. 



J 



On OX, 3 and 2 ; 


31. 


On OF, 6andl. 




(6. 2), 5 ; 


32 


On OX, 6 ± V21 ; 




On OY, imaginary points. 


33. 


(2, 4), 4Vo ; 


34. 



- On OX, and 4 ; 

lOnOZO and 8. 



29. 



30. 



40. 



(i.) C+and=Z) 2 , 
(ii.) 0+ and = E\ 
(111.) C + and > D 2 and > E 2 

x*+y 2 -10x-10y + 25^=0: 

(7, 4) and (8, 1). 

34. (2, 0) and (f, -f). 

35. fVH. 

36. ° ' - 9 - a ° 



-19: 



37. 
38. 
39. 



a' 2 + 6 2 
2x-y-2 = 0. 
4a; — 5y +71 = 0. 
3s-5v-34 = 0. 



required locus ; then the distance 



'(3, -2), 3; 
On OX, 7 and-1 ; 
On OF, ^-2± VT3. 
(-11, 9), Vl45; 
On OX, - 3 and 
-On OY, 9±2\/6. 
Let (a?, y) be any point in the 
of (a?, y) from (a^, y x ) must always be equal to its distance from (x 2 , y 2 ) ; 
therefore (x - a^) 2 + (y - y^ 2 = (a; - x 2 ) 2 + (y - y 2 ) 2 • 

whence 2x(x 1 -x i ) + 2y (y 1 - y 2 ) = {x? + y 2 - ay* - y 2 ). 

Show that this represents a straight line _L to the line joining (a^, y x ) 
and {x 2 , y 2 ). 

41. 8a? + 6y +17=0. 

42. First Method. Substitute successively the co-ordinates of the 
given points in the general equation of the circle ; this gives three equa- 
tions of condition, and by solving them we find the values of a, b, r. 

Second Method. Join (4, 0) to (0, 4) and also to (6, 4) by straight 
lines, then erect perpendiculars at the middle points of these two lines ; 
their intersection will be the centre of the circle, and the distance from 
the centre to either one of the given points will be the radius. 

.47is. x 2 + y 2 - 6x - Sy + 8 = 0. 

43. a* + y* - 8s + 6y = 0. 45. a: 2 + y 2 + 8 ax - 6 ay = 0. 

44. a; 2 + y 2 + 6x - y = 0. 46. x 2 + y 2 + 8x -f 20y + 31 = 0. 



14 ANALYTIC GEOMETRY. 

47. x 2 + y 2 -9x — 5y +14 = 0. 51. x 2 + y 2 -2ax- 2ay + a 2 = 0. 

48 fO- 5) 2 + (y + 8) 2 = 169, 52 . x 2 +y 2 = aa; + %. 

' \(x- 22) 2 + (y - 9) 2 = 169. 

^ 53. (£-l) 2 + (v-4) 2 = 20. 

• la;2 + y 2_ 6(a? -f 3/) + 9 = 0. 54 - * 2 + 2/ 2 -14a?-4 ; ?/-5 = 0. 

50. x 2 + y 2 -8x-8y + l6 = 0. 55. oj 2 + y 2 d= 2ay = 0. 

56. m{x 2 + y 2 ) + ab = (ma — &)# + (m5 — a)y. 57. x 2 + y 2 = x x x + y x y. 

58. (.t - ^) (x - x 2 ) + (y - y x ) (y - y 2 ) = 0. 6a x 2 _ ax + y 2 = r 2 _^. 

59. (1 + m 2 ) (a 2 + y 2 ) - 2 r (a; + my) = 0. 2 

Ex. 22. Page 87. 

1. The double sign corresponds to the geometric fact that two tangents 
having the same direction may always be drawn to a given circle. 
3. 2.r + 3y = 26, 3rc-2y=0, VITz, Vo2, 9, 4, yVj3. 
r 2 -^ ^ 22 (x 2 +y 2 =p 2 , 

x i " v a \yi I- (p cos a > JP s i n a )- 

5. 9 x - 13 y = 250. j- When (7 2 = r 2 (^ 2 + J5 2 ) ; 

6. z±3y = 10. 23. L f ^ Ax+By + C 

7. 104J. V.i 2 + £ 2 

8. x 2 + y 2 = 25. 24 ^ + %-"=0. 

9. 14s±6y = 232. 25 (~ a > ~ 6 )- 

10. 3z + y = 19. 26 ( 2a > &)• 

11. 3*-42/ = 0. 2 7. (0,5). 

12 f-3*-+7y.= 9&, 28 ' ^ 2 + 3/ 2 = f- 

' l3s-7y = 65. 29. m = ± f . 

13. a; = r. 30. c = 28 or - 52. 



14. ^+%±rV7l 2 +^ 2 = 0. 31. (x - 5) 2 + (y - 3) 2 - -^f. 

15. ^r-^ly±rVZ 2 T5 2 = 0. 32 J (* _ 2) 2 + (y - 4) 2 = 100, 



16. x-y± rV2 = 0. I (a? - 18) 2 + {y - 16) 2 = 100. 

17. The equation of the two tan- 33. (x -l) 2 + (y - 6) 2 = 25. 

gents is (h 2 —r 2 )y 2 =r 2 (x — h) 2 . 1 1 1 

o4. — - = — - -f- — • 

18. x + y = rv2. 



r 



19 fa = 10, 35. (a 2 +b 2 )(x 2 + y 2 )-2abVa 2 -i-b 2 

1 3 x + 4y = 50. X (* + y) + a 2 b 2 = 0. 

20. y = 2x + 13±6Vo. 36. x = a + r. 

21. 21, 3f. 37. [4r 2 -2(a-Z>) 2 ]*. 







ANSWERS. 






Ex. 23. 


Page 90. 


1. 

2. 
3. 


(1,-1), iV29. 
J(2±V7), J VII. 

(3 5\ ^/lT 




5. x 2 + y 2 = 81. 

6. (x-7) 2 +y 2 = 9. 

7. (a; + 2) 2 + (y-5) 2 = 100. 


4. 


/ % ab 


\\ . 


8. a; 2 + y 2 -2a(3a; + 4y) = 0. 



15 



^Vl + a 2 Vl + aV 
9. a; 2 + y 2 + 2& 2 + c 2 = 2 [(6 + c> + (b - c)y\ 

10. 3 ab{x 2 + y 2 ) + 2 a&(a 2 + b 2 ) = (5 a 2 + 2 6 2 )6z + (5 5 2 + 2 a 2 )ay. 

11. a; 2 + y 2 -5a:-12y=0. 13. a; 2 + y 2 + 14a; + 14y + 49 = 0. 

12. x 2 + y 2 -14a;-4y-5 = 0. 14. x 2 + y 2 - 2rx - 2ry + r 2 = 0. 

15. a; 2 + y 2 -2aa;-2ay + 3a 2 -- = 0. 

16. a; 2 -fy 2 = §. 17. 5(a; 2 + y 2 ) -10 a; + 30y+ 49 = 0. 
|( x -_ 3) 2 + (y- 1) 2 = 5, 19. 7(a; 2 + y 2 )-18a;-28y = 0. 

( \ x _ _i_t )2 + (y _ j_i)'2 = 5 . 20. a; 2 + y 2 + 50a;-f 88y-f 230 = 0. 

21 | a- 2 + y 2 - 36a; - 46y + 324 = 0, 
I 25a; 2 + 25y 2 - 80a; - 494y + 64 = 0. 

22. (6,2), 5. 25. W234. 

23. 15. 26. ViO. 

24. 10. 27. x x x + y Y y = a; x 2 + y 2 . 

28. (i.)D 2 = 4J.C, (ii.)^ 2 = 4^C, (Hi.) D 2 = E 2 = ±AC 

29. r 2 = 2rmc + c 2 . 35. 6x -8y- 25 = 0. 

30. £ = 40. 36. 2(37±3\/41)a;-f 25y=0. 

31. a; 2 +y 2 = ayV2. 37. x + VSy± 20 = 0. 

32. a 2 +y 2 ±2a(a?±y) = 0. 38. a;+y-10 = 0. 

33. 2(a; 2 - ax + y 2 - r 2 ) + a 2 = 0. 40. 3(35 + 24V30). 

34. a;-y = 0. 43. 135°. 
C (x + 4) 2 + (y + 10) 2 = 85. 44. (7, - 5) and (- 6^, 9ff). 



45. 



-j / 514V , / , 670V 85 
U"-i69J + ( y+ r69J = T69 2 ' 



46. The circle (x — oci) 2 + (y — y x ) 2 = r 2 . 

47. The circle (x - a) 2 + (y - b) 2 = (r + r') 2 . 

48. The circle (x - a) 2 + (y - 6) 2 = r 2 + £ 2 . 



16 ANALYTIC GEOMETRY. 

49. Take A as origin, and let the radius of the circle = r ; then the 
locus is the circle x 2 + y 2 = rx. 

50. Take A as origin, and let the radius of the circle = r ; then the 

i • u, • i 2 , 2 2mrx 
locus is the circle x* + y z = 

° in + n 

51. Take A as origin, JLi? as axis of a, and let J. 5 = a ; then the 
locus is the circle (m 2 + n 2 ) {x 2 + y 2 ) — 2 am 2 x + a 2 m 2 = 0. 

52. Take AB as the axis of x, the middle point of AB as origin, and 

let AB = 2a ; then the locus is the circle 2(x 2 + ?/ 2 ) = k 2 —2a 2 . 

53. Using the same notation as in No. 52, the locus is the straight line 
4aa? = k 2 . 

54. Taking the fixed lines as axes, the locus is the circle 4(# 2 + y 2 ) = d 1 . 

55. Take the base as axes of x, its middle point as origin, and let the 
length of the base = 2a, and the constant angle at the vertex = 6. Then 
the locus is the circle x 2 + y 2 — 2 a cot 6 y = a 2 . 

56. Take A as origin, AB as axis of x, and let AB = a, AC=b. 

b 2 
Then the locus is the circle (x — a) 2 + y 2 = — • 

4r* 

57. The circle x 2 + y 2 = where £ is the length of the chord. 

4 r 2 — I 2 

58. The locus consists of the two circles x 2 + y 2 ± ro = 0. 



Ex. 24. Page 103. 

1. 7x-6y = 0. 2. 2z-2y + 9 = 0. 

4. a: + y = r, 2 a? + 3 y = r, (a + b) x + (a — 5) y = r 2 . 

5. 13z + 2y = 49. 6. The tangent at (M). 

7. (i.) 2» + 3y = 4, (ii.) 3z-y = 4, (iii.) x - y - 4. 

8. (i.) (20, 30), (ii.) (21, -14), (iii.) (35 a, 35 ft). 

9. (6, 8). 18. V2x + 11 y- 51 = 0. 
10. [-^!, -^- 2 \ 19. x + y-2 = 0. 

ii. Lv^),Lv:iT^ = 4. 20 («'-«»)*-( «»-^ + <*- o- 

16. h 2 + k 2 - r 2 . 21 * - y- °. ^(a + ft) 2 - 4 c. 

17. 3. 22. (-2,-1). 






ANSWERS. 17 



Ex. 25. Page 115. 

1. Writing x + 1 for a?, and y — 2 for y, and reducing, we have y 2 = 4#. 

2. x 2 + y 2 = r 2 . 5. a; 2 + y 2 = r 2 . 

3. o; 2 + y 2 = 2ra;. 6. 2a;y = a 2 . 

4. z 2 -fy 2 = -2n/. 7. a 2 - y' 2 = 2. 

8. (i.)p = o, (ii.) p 2 cos 2 = a 2 . 

9. (i.) p = 4 a cot cos 0, (ii.) p(l — cos 0) = 2 a. 

10. (i.) x 2 -f y 2 = a 2 , (ii.) a; 2 + y 2 = ax, (iii.) a; 2 — y 2 = a 2 . 

11. a; + y = 0. 15. a; 2 — 6 a^/ + y 2 == 0. 

12. 2x-5y = 0. 16. a:y = 3. 

13. 12;r 2 +16a:y + 4y 2 = l. 17. y 2 = 2a(a;V2 - a). 

14. z 2 + y 2 = 25. .18. 4zy==25. 



Ex. 26. Page 117. 

1. &V3. 12. 9a; 2 + 25y 2 = 225. . 

2. 4 sin J co. 13. /> = 8acos0. 

3. Vl3-12cosco. 14. p = 4a. 

4. Va 2 + b 2 - 2 ah cos (0 - c/>). 15. /> = fcsc 2 j0. 

5. 2 a sin0. 16. p = 49 sec 20. 

6. 2a cos 0. 17. /> = Fcos20. 

7. aV5 - 2V3. 18. a;y = a 2 . 

9. 2z 2 + 2a:y +y 2 =l. 19. (a; 2 + y 2 )* = 2 kxy. 

10. 2a; 2 + y 2 = 6. 20. ar 5 + y 3 - 5 kxy = 0. 

11. y = 0. 21. tan- 1 !. 

22. (i.)tan- 1 f--\ (ii.)tan- 1 -- 

Ex. 28. Page 124. 

2. y 2 = ipx-4:p 2 . 3. y 2 = 4p:r + 4p 2 . 

4. (i.)y 2 = 10.T, (ii.)y 2 = 10aj + 25, (iii.) y 2 = 10 x -25. 

5. (i.) y 2 =l6x, (ii.) 3/ 2 =16^ + 64, (iii.)y 2 =16a;-64. 



18 ANALYTIC GEOMETRY. 

6. (2, ± 6). 8. (4, 6) and (25, 15). 

7. 6, 15, -• 9. (12,6). 

b 

10. The line a? = 9 meets the parabola in (9, 6) and (9, -6). The 
line a? = passes through the vertex. The line x = — 2 meets the para- 
bola in the imaginary points (— 2, ± V— 8). 

11. The line y = 6 meets the parabola in (9, 6). The line y = — 8 
meets the parabola in (16, — 8). 

12. p = 4. 13. The point (2, 8). 

14. (i.) y = 0, (ii.) ».== - 2, (iii.) a? = 2, (iv.) a; ± y - 2 = 0, (v.) y = 2a?. 

15. (i.) 4* -5?/ + 24 = 0, (ii.) a; 2 + 2/ 2 = 20a?. 16. 3p. 17. 8V3p. 

26. The common latus rectum = 4p. The common vertex is at the 
origin. The axis of a: is the axis of (i.) and (ii.) ; that of y is the axis of 
(iii.) and (iv.). Parabola (i.) lies wholly to the right of the origin, 
(ii.) wholly to the left, (iii.) wholly above, (iv.) wholly below. We may 
name them as follows : — 

(i.) is a right-handed X-parabola. 
(ii.) is a left-handed X-parabola. 
(iii.) is an upward F-parabola. 
(iv.) is a downward Y"-parabola. 

27. (i.) If J. = 0, the equation becomes y' 2 + By + (7=0, a quadratic 
in y, and representing the two straight lines, parallel to the axis of x, 

r> 

whose equations are y ± — ± J V5 2 — 4 A (7= 0. (ii.) If B = 9, the origin 

is in the axis of the curve, (iii.) If (7=0, the origin is in the curve, 
(iv.) If A = 5 = 0, the locus consists of the two parallels y = ± V— C. 
(v.) If A = C = 0, the locus consists of the straight lines y = 0, y + i? = 0. 
(vi.) If B= C= 0, the vertex is at the origin. 

28. Latus rectum = — B. 
The vertex is the point ' 

\ f *±J3 j 

The focus is the point ( - ^< ^ 7^° ~ t\ 

The axis is the line x = 

The directrix is the line y = 

* 45 

If B is negative, the curve is above the axis of a?. 

If B is positive, the curve is below the axis of x. 






2 


45 


; 


A 

A 


A 2 - 4 C 
45 


_5 


2* 
A 


2 + 5 2 - 


4(7 



ANSWEES. 



19 





Latus 
Rectum. 


Vertex. 


JFbews. 


Axis. 


Directrix. 


29. 


12 


(7,0) 


(10, 0) 


y = 


X = 4: 


30. 


12 


(- 7, 0) 


(-4,0) 


y = 


£ = -10 


31. 


-12 


(- 7, 0) 


(- 10, 0) 


y— o 


£ = -4 


32. 


-12 


(7,0) 


(4,0) 


y = o 


a; = 10 


33. 


12 


(0,7) 


(0, 10) 


a; = 


y = 4 


34. 


8 


(6,4) 


(3,4) 


y = 4 


x = 4 


35. 


~f 


(-i.o) 


(-1,0) 


y = 


* = -i 


36. 


1 


».-.» 


(J. -2) 


» = £ 


2/ = -f 


37. 


4 


(- 2, - 3) 


(-1,-3) 


y — 3 


£ = — 3 



Ex. 29. Page 129. 

2. On the axis of x lay off from the vertex to the left a length equal 
to the abscissa of the point of contact ; this determines a second point in 
the tangent. Lay off from the foot of the ordinate of the given point, 
towards the right, a length equal to 2p ; this determines a second point 
in the normal. 

6. £-4y + 20 = 0, 4:X + y-90 = 0. 

*-2/ + 3 = °> normals j ^ + 2/ - 9 = 0, 

These lines enclose a square whose area = 72. 

8. Tangent = V266, normal = \A)5, subtangent = 14, subnormal = 5. 

P 



7. Tangents 



9. (5, 10). 



13. 



14. [^ 



v 2 > i(yi+yj)]- 

2j)), intercept = p. 



iVl 4- m 2 

15. # — y +_p = 0, point of contact (p. 

16. Equation of the tangents y V3 = ± £ ± 3_p, required point (— 3p, 0). 

17. For the two points whose co-ordinates are 

S = f(l ± Vl7), y- ± ^l±^I 

18. For the points (0, 0) and (3p, ± 2p VS). 

19. 9z-6y + 5 = 0, (f, f). 

20. a?-2y +12 = 0, (12, 12). 



20 ANALYTIC GEOMETRY. 



21. 2/ = z(±V2-l) + 4(±V2-l). 22. ^ + ,T ^ . 

24. One of the points of contact is (— 1, 11). The vertex of the para- 
bola is the point (— 9, 3) ; therefore the distance from the vertex to the 
intersection of the axis and the ordinate of the point of contact is 
eqnal to 8 ; therefore the subtangent = 16 ; therefore a second point of 
the tangent is (—17, 3) ; therefore the equation of the tangent through 
(-1, 11) is x-2y + 23 = 0. 

25. j (ii.) x Y x= 2p(y + y 1 \ 
t (iii.) x 1 x = -2p(y + y 1 ). 



Ex. 30. Page 131. 

1. y 2 = 24z-144. 2. y 2 =16x. 3. y 2 = -l7x. 

4 f 2y 2 — IKt +122/ + 73 = ; or 5. (y + 7) 2 = 4 (a- 3). 

' 12* 2 + 11* +122/ -37 = 0. 6 3y 2 = ±x. 

7. 3a; 2 = 4y. 8. §, 8z + 3 = 0, 8z±15y-3 = 0. 

1Q | 4 on OX; 20. y 2 = 9x. 



8 and - 2 on O F. 



21. y 2 = l 



11. 4(2 + V3)p. 4r 2 -^ 2 

22. y 2 = x. 

r 00 y =_* + 2 > r 

12. (ii.) 2V2, 23 f = t x , 

C (iii.) x + y -6 = 0. r- 

13. x + y- 6 = 0. 24. y 2 =- 



V n 2 + * 2 



2, 10 

16 (2 4) (11 10) 27. The equation of the circle is 



14- y-yi = -^( x ~ x i)- 25. 2/ 2 =2(2r-s)z. 

15. (8, 4), (2, 10). 26. 4 P^- 



17. y = 



b ± Vb 2 — 4 op 



2a 

[ A left-handed X-parabola. 
I Latus rectum = — 2. 

18. \ Vertex. (-2, 0). 
j Focus, (- |, 0). 

[ Directrix, a?=« — §. 

19. 4 a. 





(x-3) 2 + (2/-f) 2 = 


28. 


(-3p,0). 


29. 


(?*& 


SO 


1 45° and 135°. 




31. 


4^ 2 . 


34. 


The parabola y 2 =px 



ANSWERS. 21 

The loci in exercises 35-38 are parabolas, the latus rectum in each 
being half that of the given parabola. If the given parabola is y 2 = 4p#, 
the equations of the loci are : 

35. y 2 = 2px-p 2 . 39 ' TLe strai S ht line V = P h 

36. y 2 = 2px - 2p 2 . 40 Tlie P arabola 3/ 2 - ±P X =jM 2 - 

^ o 41. The straight line Jcx — p. 

37. y 2 = 2px. & ^ 2 

38. y 2 = 2px + 2p 2 . 42 The parabola (a;-^?/ 2 ^- 

43. Take the given line as the axis of y, and a perpendicular 
through the given point as the axis of x, and let the distance from the 
point to the line = a. The locus is the parabola y 2 = 2al x — - )• 

Ex. 31. Page 143. 

10. 3z-5y-6 = 0. [2ay = 4:px. 

1 1 8 _ 25 — 14 J ^ be cnor d is parallel to the 

I tangent at the end of the 

12. I3x + 22y + k = 0. I diameter. 

13. x -y -1=0. 15. y 2 = 52x. 

20. If the equation of the given parabola is y 2 = 4p#, the locus is 
the parabola y 2 = p(x —p). 

21. If the equation of the given parabola is y 2 = 4px, the locus is 
the parabola y 2 =p (x — 'dp). 

22. Take the given line as the axis of y, and a perpendicular through 
the centre of the given circle as the axis of x. Let the radius of the 
circle = r ; distance from the centre to the given line = a. There are 
two cases to consider, since the circles may touch the given circle either 
externally or internally. The two loci are the parabolas 

y 2 = 2 (a + r) x + r 2 — a 2 , 
y 2 = 2 (a — r) x -f r 2 — a 2 . 

23. Let 2a be the given base, ah the given area; take the base as 
axis of x, its middle point as origin ; then the locus is the parabola 

x 2 + by = a 2 . 

Ex. 32. Page 153. 
1. 5, 4, 3, f. 2. V2, 1, 1, Vl 3. 2, V3, V7, JV7. 



4. 



1 1 J A-B 1 J A-B 
V3' VB A ' ^b * A\ B 



22 ANALYTIC GEOMETRY. 

5. fV6. 13. £ 2 -f 2y 2 = 100. 

6. e = jV3. 14. 8z 2 -f 9y 2 =8a 2 . 

7. 4z 2 + 9y 2 = 144. 15. 2 : V3. 

8. 25a> 2 +169y 2 = 4225. 1ft a& 

■* 16. # = y = ± — . 

9. 144a; 2 + 225y 2 = 32,400. Va 2 + b 2 

10. 16z 2 + 25y 2 = 1600. 18. (if), (-|,-f). 

11. 25* 2 +169y 2 = 4225. 19. (1, 2), (1, - 2). 

12. 115 a: 2 + 252y 2 = 4140. 20. (3, 1), (3, -1), (-3, 1), (-3, -1). 

21. See No. 11. The equation of the locus is x 2 + 2y 2 = r 2 . 

22. Taking as axes the two fixed lines, and putting AP=a, BP= b, 
the acute angle between AB and the axis of a? = <£, we find that 

x = a cos (p, y = b sin (p. 
Therefore P describes an ellipse whose equation is 






2 7,2 



ar 

a 2 ' 5 2 

23. The two straight lines y = ± — x. The locus is imaginary when 

B 

the values of y are imaginary ; that is, when A and i? have unlike signs. 
29. The equations of the sides are 

ab ab 

y = - 



Va 2 + b 2 Va 2 + 6 2 

4a 2 5 2 

area = 

a 2 + 6 2 



Ex. 33. Page 160. 



1. 



4a;±9y = 35, 7. 2y = z + 10. 

9z + 4y=6, 8. 4z.-3y± Vl07 = 0. 



i 2a;±3yV3-12 = 0, 
A 1 *__/^. *_„ , n.^j , o./5 a 9. x= , y 



6o?\/3±4y+9V3+2V6=-0. 

x + y = 10, 
rc-4y + 6 = 0; 



Va 2 +7/ 2 Va 2 + & 2 

^ ~~ ' 10. Same answers as No. 9. 





1 8, J. 




11. b l : a z . 


5. 


m 2 ?i 2 




12. a; = ±- a -, y=±A. 
V2 V2 


6. 


9a; 2 + 25y 2 = 


225. 


13. y = 4, 3z + 34y=17. 



ANSWERS. 23 

14. The equation ± Vdx ± oy = 9 a represents the four tangents. 

15. aVl — e 2 cos 2 cf>. 16. J (a 2 tan ^ + b 2 cot <£>). 

17. The extremities of the latera recta. 

19. The method of solving this question is similar to that employed 
in $ 147. The required locus is the auxiliary circle x 2 + y 2 = a 2 . 

Ex. 34. Page 161. 

r £=*8, 40y = 9z + 72; 8. kc + ay = a& V2. 

55 9 - cos d> + y sin d> = 1. 

2. Within. a 7 & 

x 10. a5 ■ 

3. — —• Va 2 — e 2 x-. 2 
V3 



]_ 11. aVl — e 2 cos 2 <£. 

4 * Vi' 12. e 2 :& 2 . 

5 VI3+1 13 ^M.i. 



2V3 



a 2 fr 



2A2 



e 



6. a: ± y ± Va 2 + fr 2 = 0. 14 - V(l- e 2 ) (a 2 - eV). 

7. ox + cy = bc\ — ■ 15. tan <p = 

u * a 2 + b 2 Y e 

18. The locus is the minor axis produced. 

19. The ellipse [ x + — = — ; centre is ( -. ] ; semiaxes are - 

andr . I 2) ± 4' ^ U )' 2 

20. The ellipse a 2 1 y ] + b 2 x 2 = : centre is ( 0, - ] ; semiaxes 

a ,h \ 2 > 4 V 1) 

are - and — 

2 2 

In 21-23 take the base of the triangle as the axis of x, and the origin 
at its middle point. 

21. The ellipse (s 2 - c 2 ) x 2 + s 2 y 2 = s\s 2 - c 2 ). 

22. The ellipse kx 2 + y 2 = he 2 . 

23. The circle + cf + y 2 = 4 a 2 . 

Ex. 35. Page 176. 

1. x = ±- = ±-. 3. 20a; + 63?y-36 = 0. 

c e 

Ao? Bb 2 \ Q r v o b 2 r . x 9 5 .... N „ - 

~7T l "Try a ( l -) m =-? (n.)™ 2 = -' (m.) m 2 = l. 
o o y a" 5 a 



24 ANALYTIC GEOMETRY. 



-w^i 



9. a = l\ L ~ e C ° b * , 6 = ZVl-e 2 cos 2 0. 



- t* 
10. 3^4-8y = 4, 2z-3y = 0. 11. a 2 y l x = b 2 x 1 y. 

yi 

12. Area = - — (m 4- n), m and 71 being the two segments (use the 

A a 
polar equation). 

13. 26z + 33y-92 = 0. 14. z-f2y = 8. 
15. b 2 x + a 2 y == 0, 6 2 a? — a 2 y = 0, a 3 y + 6 3 # = 0. 



16. 


6:r + ay = 0. 


17. 


See | 157. 


18. 


See g 158. 


23. 


a 


24. 


y\/a 2 -l 2 = 


25. 


e = aV6. 



26. 


See | 156. 


27. 


See § 159. 


30. 


X 2 T/ 2 _ 2 a: 
a 2 6 2 a 


31 


«(!-*) 




1 — e cos 



32. ,~ h2 



1— e 2 cos 2 

34. 16 a; 2 + 49 y 2 -128 b - 6863/ + 1473 = 0. 

35. 2a = 18, 26 = 10. 36. t±l^ + y 2 = 5 a. 

144 J 

37. Centre is (-1, 1), 2a = 2, 26 = 4. 

38. Centre is (2, 4), 2a = 8, 25 = 6. 

39. Centre is (0, - J), 2a = 3, 26 = 1. 

40. cos <p = AP^- - 41. tan^ = -^=A. 

N a 2 -6 2 + Vo6 

42. Find the ratio of y' to the intercept on the axis of y. 

43. Vh» + a% = BW + *•*•. 47 The em £ 2 + ^ = J. 

,2 ,,2 a 2 6 2 

46. The ellipse — + £- = 2. 

r a 2 6 2 48. The ellipse 6V + a 2 y 2 = 6 2 c 2 . 

49. The ellipse 25 z 2 + 16 y 2 - 48 y = 64. 







Ex. 


36. 


Page 186. 


1. 


x 2 y 2 _ -I 

64 49 "" 

25 144 






3. 8x 2 -y 2 = 8a 2 . 


2. 






4. 625 x 2 - 84 y 2 = 10,000. 

5. 2£ 2 -2t/ 2 = c 2 . 



7. a = 4, 6 = 3, c = 5, e = J, latus rectum = f . 



in L 

Lb ua u-- 



ANSWERS. 25 

8. lQy 2 — 9x 2 = 25, transverse axis = 6, conjugate axis = 8, distance 
between focus = 10, latus rectum — - 3 3 2 -. 

9. a:b = l: V3. 11. e=V2. 

14. Foci, (5, 0), (- 5, 0) ; asymptotes, y = ± ±x. 17. b. 

Ex. 37. Page 188. 

1. 16z-9y = 28, 9z + 16y = 100, f, - 6 ^. 3. x 2 -y 2 = 9, (5,4). 

4. The four points represented by 

± a 2 ± 6 2 
# = — ., y = 

Va 2 - 6 2 Va 2 - B* 

If 5 2 > a 2 , the points are imaginary. If the hyperbola is equilateral, 
the points are at an infinite distance. 

n 2 n 2 7,2 

9. — • 10. — ---1. 

^/3 ?ti 2 n 2 

11. When a is less than b. 12. The circle x 2 + y 2 = a 2 . 

Ex. 38. Page 189. 

1. 25e, ae 2 . 11. y = ±a;\/2 + a. 

2. 14 and 6. 12. (0, ± Va 2 - b 2 ). 

3. The sum = 2 ex. 13. Z> 2 > a 2 . 

8. (a, 6 V2), (a, - &V2), 14. 64 a? - 9y + 741 = 0. 

10. They are equal. 15. y = 4 a ± 8 V2. 

/7 2 7) 2 

16. x=± a - , y = :" 



Va 2 - 4 6 2 Va 2 -46 2 

Ex. 39. Page 201. 
1. 9z + 12y + 16«0. 6. a. 

2 a; = ±^. 8 75.T-16y=0. 

e 9. 245 a: -12y- 1189 = 0. 

3. * 

2 

4 /_ ^. 2 BW\ 
5. x -f- a = 0. 



10. 


|V3. 


17. 


See \ 150. 


18. 


4a:y=-(a 2 + 5 2 ). 


19. 


a; y 



26 ANALYTIC GEOMETRY. 



v ' ; a 2 b 2 a ' l-ecos0 

.(«•) ^-r»+-r =a 23. P 2 = 



a 2 6 2 a \ r e 2 cos 2 0-1 

24. The equation represents an equilateral hyperbola, with its trans- 
verse (real) axis parallel to the axis of y. The centre is (1, — 2) ; the 
semiaxes are each equal to 2. 

25. The hyperbola 3 a; 2 ,- y 2 + 20a:- 100 = 0. The centre is the point 
(— ™, 0). Changing the origin to the centre, we obtain 9x 2 — 3y 2 = 400. 

26. The locus is the curve 2xy — 7x + 4y = 0. If we change the 
origin to the point (h, k), we can so choose the values of h and k as to 
get rid of the terms containing x and y. Making the change, we obtain 

2xy + (2 k - 7) x + (2 h + 4)^ - th + 4>h •+ M •-* 0. 
If we choose A and & so that 2A + 4 = 0, and 2 A: — 7=0, that is, if 
we take A = — 2, & = |, the terms containing a; and y vanish, and the 
equation becomes 2xy = 7. Hence we see ($ 194) that the locus is an 
equilateral hyperbola, and that the axes of co-ordinates are now the 
asymptotes. 

27. The equilateral hyperbola 2a;y = a 2 . 

28. Taking the base as axis of a;, and the vertex of the smaller angle 
as origin, the loci are the axis of x and the hyperbola 3x 2 — y 2 — 2ax = 0. 

Ex. 40. Page 221. 

1. The ellipse 72a? + 48 y 2 = 35. 9. The ellipse 4 a; 2 + 9y 2 = 36. 

2. The parabola y 2 = — fa?. 10. The parabola y 2 = 2 x. 

3. The parabola y 2 =2x^2. 11. The ellipse 4ai 2 + 9?/ 2 = 36. 

4. The ellipse 4a; 2 + 2y 2 = l. 12. The ellipse 4a; 2 + ?/ 2 = 100. 

5. The hyperbola 32a: 2 -48y 2 = 9. 13. The hyperbola 4a; 2 - 9y 2 = 36. 

6. The parabola y 2 = 3a; V'2. 14. The straight lines y=x, y= — 5. 

7. The ellipse 9a; 2 + 3/ = 32. 15. The parabola y 2 = ox. 

8. The hyperbola 4a; 2 -4t/ 2 +l=0. 16. The parabola 25x 2 +2yV5 = 0. 



